# Person A travels up in an elevator at uniform acceleration. During the ride, he drops a ball while Person B shoots an arrow upwards directly at the ball. How much time will pass after Person B shot the arrow before the arrow hits the ball?

## Person A gets into a construction elevator (it has open sides) at ground level. He is carrying a Styrofoam ball. The elevator starts to travel upwards, accelerating uniformly at a rate of $1.2 \frac{m}{s} ^ 2$. After the elevator has been moving $8.0 s$, Person A drops the ball over the side of the elevator. The Styrofoam ball, being very light, accelerates downwards at a rate of $3.8 \frac{m}{s} ^ 2$. Person B is standing on the ground with a bow and arrow. At the instant when Person A drops the Styrofoam ball, Person B shoots an arrow upwards at a speed of $32 \frac{m}{s}$ directly at the ball. How much time will pass after Person B shot the arrow before the arrow hits the ball?

Mar 23, 2016

This solution is not really valid. Please see the other solutions which are better.

$t = 1.378 s$. If we assume that the ball starts at zero velocity when it is dropped, explained in Phase 2 of the solution.

#### Explanation:

I will consider the problem in two phases. The first phase is the motion of the elevator before the ball is dropped, the second phase is after the ball is dropped and the arrow is shot upward.

Phase 1: Elevator accelerating upwards.
Acceleration is constant so we can use an equation of constant acceleration to determine the height, h, from which the ball will be dropped.
$s = h$
$u = 0$
v = ?
$a = 1.2 m . {s}^{- 2}$
$t = 8.0 s$

Use this equation:
s = ut + ½at² => h = 0 + 0.5 × 1.2 × (8.0)² = 38.4 m

Phase 2: Ball dropped from elevator.
In this solution I will assume that the ball is dropped with zero initial velocity. However, because the elevator has an upward velocity of $9.6 m . {s}^{- 1}$ the ball actually would also start with that velocity. The reason that I will assume it is dropped with zero initial velocity is that it simplifies the solution significantly; or put another way without that assumption the solution is much more difficult.

Furthermore, I believe that the question implies we should make that assumption because it states that the ball "accelerates downwards with acceleration of $3.8 m . {s}^{- 2}$". Whereas if it started moving up then it would decelerate upwards with at least $9.8 m . {s}^{- 2}$.

So after $8.0 s$ the ball is dropped from $38.4 m$ and the arrow is shot upwards at $32 m . {s}^{- 1}$. After time, t, they collide. If we define the distance the ball drops until the collision as y then the situation is as shown in the diagram below. It also follows that in time t the arrow will travel a distance of h-y.

Now apply the equations of constant acceleration to the ball, then to the arrow and then use simultaneous equations to solve for t. In both cases we will use the equation: s = ut + ½at²

Ball
${s}_{b} = y$
${u}_{b} = 0$
v_b = ?
${a}_{b} = 3.8 m . {s}^{- 2}$
${t}_{b} = t$

s = ut + ½at² ⇒ y = 0 × t + ½ × 3.8 × t² = 1.9 t²
Equation ①: y = 1.9 t²

Arrow
${s}_{a} = h - y$
${u}_{a} = 32 m . {s}^{- 1}$
v_a = ?
a_a = –9.8 m.s^(-2)
${t}_{a} = t$

s = ut + ½at² ⇒ h – y = 32 t + ½ × (-9.8) × t² = 32 t – 4.9 t²
Equation ②: ⇒ y = 38.4 – 32 t + 4.9 t²

Equation ① = Equation ②:
1.9 t^2 = 38.4 – 32 t + 4.9 t^2
⇒ 3.0 t^2 – 32 t + 38.4 = 0
⇒ t^2 – 10.67 t + 12.8 = 0

Factorise the quadratic to find solutions for t:
(t – 9.292)(t – 1.378) = 0
⇒ t_1 = 1.378 s
⇒ t_2 = 9.292 s
The solution that we want for this problem is ${t}_{1}$ because the longer time is the time at which the arrow and ball would collide a second time (if that were possible).

So the solution is $t = 1.378 s$.

We can check this solution by passing the value of t back into equations ① and ②.

Equation ① : y = 1.9 × (1.378)^2 ⇒ y = 3.608 m
Equation ② : h – y = 32 × (1.378) – 4.9 × (1.378)^2 = 44.10 – 9.305 = 34.791 ⇒ y = 38.4 – 34.791 = 3.609 m

As you can see the two values for y are consistent, so the value of t should be accepted.

Mar 24, 2016

Without assuming that the ball starts with zero initial velocity the time taken would be:
$t = 1.51 s$

#### Explanation:

Plot spoiler: I do not assume that the ball is released with zero initial velocity in this solution. This is a long solution with some fairly complex assumptions, it is not for the faint hearted!

I will consider the problem in three parts. The first part is the motion of the elevator before the ball is released, the second part is between the ball being released and reaching its maximum height, and the third part is between the ball starting to fall downwards and the arrow colliding with the ball.

Part 1: Elevator accelerating upwards.
Acceleration is constant so we can use an equation of constant acceleration to determine the height, h, at which the ball will be released.
s = h = ?
$u = 0$
v = ?
$a = 1.2 m . {s}^{- 2}$
$t = 8.0 s$

Use this equation:
s = ut + ½at² => h = 0 + 0.5 × 1.2 × (8.0)² = 38.4 m

We also need to know the velocity of the elevator at this height as the ball will have this as its initial velocity:
v = u + at = 0 + 1.2 × 8.0 = 9.6 m.s^(-1)

Part 2: Ball released from elevator.
The ball is released with an upward velocity of $9.6 m . {s}^{- 1}$. It will decelerate, reach a maximum height and then start to accelerate downwards. In part 2 we will calculate the maximum height the ball reaches and the time it took to get there. We will also calculate how far the arrow travels in this time and what its final velocity is.

The situation is shown in the diagram below. ${d}_{b}$ and ${d}_{a}$ are the distances travelled by the ball and arrow respectively. x is the distance between ball and arrow when the ball is at maximum height.

An important note about how I have treated drag in this solution.
The question does not give us sufficient information to correctly handle drag in this question. Drag is a function of velocity squared, so the drag in reality would increase as the ball accelerated and vice versa. But the question gives us a fixed value of the acceleration of the ball whilst it is moving downwards ($3.8 m {s}^{- 2}$); we do not know the function of the drag. Furthermore, the question apparently fails to take into account that the ball initially moves upwards when it is released.

So I have made the following assumptions in order to write something that gets as close as possible to a proper solution:
1. The drag does not change as a function of velocity squared.
2. The value of the acceleration due to drag is constant in all cases. When the ball is going down drag changes the acceleration from $9.8 m {s}^{- 2}$ to $3.8 m {s}^{- 2}$. That is a change of $6.0 m {s}^{- 2}$. So the acceleration due to drag is $6.0 m {s}^{- 2}$.
3. The ball does not reach terminal velocity in either aspect of its motion.

Height of the Ball and Time of Travel:
If you notice in the diagram I drew the forces acting on the ball. Whilst it is travelling upwards drag and weight act downwards. So the accelerations due to them both will be added together to find the resultant acceleration. Noting the above assumptions the upward deceleration is ${a}_{b} = 9.8 + 6.0 = 15.8 m . {s}^{- 2}$.

s_b = d_b = ?
${u}_{b} = 9.6 m . {s}^{- 1}$
${v}_{b} = 0$
a_b = –15.8 m.s^(-2)
t_b = ?

For the height use this equation: v² = u² + 2as ⇒ s = (v² - u²)/2a
⇒ d_b = (0 – 9.6²)/(2 × (–15.8)) = 0.3038 m
For the time of travel use this equation: v = u +at ⇒ t = (v – u)/a
⇒ t = (0 – 9.6) / -15.8 = 0.6076 s
Don’t forget to add this time to what is calculated in part 3.

How far the arrow travelled during this time and its final velocity:
s_a = d_a = ?
${u}_{a} = 32 m . {s}^{- 1}$
v_a = ?
a_a = –9.8 m.s^(-2)
${t}_{a} = 0.6076 s$

For the height use s = ut + ½at²
⇒ d_a = 32 × 0.6076 + 0.5 × (-9.8) × 0.6076² = 17.63 m
For the final velocity use $v = u + a t$
⇒ v_a = 32 + (–9.8) × 0.6076 = 26.05 m.s^(-1)

So when the ball reaches maximum height the distance between ball and arrow, x, is:
x = h + d_b – d_a = 38.4 + 0.3038 – 17.63 = 21.07 m

Part 3: From ball starting to drop downwards to collision.
The situation now is as shown in the diagram below. The total distance between ball and arrow is x and the ball falls through distance y before colliding with the arrow. So the arrow therefore moves through distance x – y before colliding with the ball.

Now apply the equations of constant acceleration to the ball, then to the arrow and then use simultaneous equations to solve for t. In both cases we will use the equation: s = ut + ½at²

Ball
${s}_{b} = y$
${u}_{b} = 0$
v_b = ?
${a}_{b} = 3.8 m . {s}^{- 2}$
${t}_{b} = t$

s = ut + ½at² ⇒ y = 0 × t + ½ × 3.8 × t² = 1.9 t²
Equation ①: y = 1.9 t²

Arrow
${s}_{a} = x - y$
${u}_{a} = 26.05 m . {s}^{- 1}$
v_a = ?
a_a = –9.8 m.s^(-2)
${t}_{a} = t$

s = ut + ½at² ⇒ x – y = 26.05 t + ½ × (-9.8) × t² = 26.05 t – 4.9 t^2
Equation ②: ⇒ x – y = 26.05 t – 4.9 t^2

Substitute for y in equation ②:
x – 1.9 t^2 = 26.05 t – 4.9 t^2
$x = 21.07 m$
⇒ 3.0 t^2 – 26.05 t + 21.07 = 0
⇒ t^2 – 8.683 t + 7.023 = 0

Factorise the quadratic to find solutions for t:
(t – 0.903)(t – 7.78) = 0
⇒ t_1 = 0.903 s
⇒ t_2 = 7.78 s
The solution that we want for this problem is ${t}_{1}$ because the longer time is the time at which the arrow and ball would collide a second time (if that were possible).

So our solution is $t = 0.903 s$.

Now add to that the time calculated in part 2 to give the final solution:
$t = 0.903 + 0.6076 = 1.51 s$

We can check the quadratic solutions by passing the value of t back into equations ① and ②.

Equation ① : y = 1.9 × (0.903)^2 ⇒ y = 1.549 m
Equation ② : y = x – 26.05 × (0.903) + 4.9 × (0.903)^2 = 21.07 – 23.52 + 3.996 ⇒ y = 1.546 m

As you can see the two values for y are consistent, so the value of t should be accepted.

Apr 5, 2016

8/3s

#### Explanation:

The elevator starts with initial velocity Zero and with acceleration $1.2 \frac{m}{s} ^ 2$ .After 8s it will reach at the height,$h = 0 \times 8 + 0.5 \times 1.2 \times {8}^{2} = 38.4 m$ and
also attains velocity,$v = 0 \times 8 + 1.2 \times 8 = 9.6 \frac{m}{s}$
At this moment (just completion of 8s) the person A drops the ball and person B shoots the arrow from the ground with initial upward velocity,$32 \frac{m}{s}$ and having downward acceleration $9.8 \frac{m}{s} ^ 2$ to hit the ball which has downward acceleration $3.8 \frac{m}{s} ^ 2$

Let after $t s$ of its shoot the arrow hits the ball.
During this ts if arrow ascends height ${h}_{1} m$
Then ${h}_{1} = 32 \times t - \frac{1}{2} \times 9.8 \times {t}^{2.} \ldots \ldots \ldots \left(1\right)$

Again during this t s if the ball ball ascend ${h}_{2} m$ from the point it was dropped, then ${h}_{2} = 9.6 \times t - \frac{1}{2} \times 3.8 \times {t}^{2.} \ldots \ldots \ldots . . \left(2\right)$

Obviously ${h}_{1} - {h}_{2} = 38.4 m$ is the initial distance of separation between the ball and arrow at the moment the ball was dropped or the arrow was shot.
So subtracting Eq (2) from Eq (1) we can write
$\left(32 - 9.6\right) \times t - \frac{1}{2} \times \left(9.8 - 3.8\right) \times {t}^{2} = {h}_{1} - {h}_{2} = 38.4$
$\implies 22.4 t - 3 {t}^{2} = 38.4$
$\implies 3 {t}^{2} - 22.4 t + 38.4 = 0$
$\implies t = \frac{22.4 \pm \sqrt{{\left(22.4\right)}^{2} - 4 \times 3 \times 38.4}}{2 \times 3} = \frac{22.4 \pm 6.4}{6}$
$\therefore t = \frac{8}{3} s \mathmr{and} 4.8 s$

smallest value of t$= \frac{8}{3} s$ is the time required by the arrow to hit the ball.
If the arrow bypasses the ball without hitting then second meeting is possible and the second value of t = 4.8 s is the time of second crossing when both ball and arrow move downward in the back journey

Apr 5, 2016

To add to existing solutions, here is one more.

$\approx 2.63 s$

#### Explanation:

The problem is dealt in two time-phases.

• The person with Styrofoam ball travels up in the elevator.
Elevator equation. Let us take up as $+ v e$ direction. Given that
$u = 0$
$a = 1.2 m . {s}^{- 2}$
$t = 8.0 s$

We need to ascertain what was the velocity ${u}_{b}$ and height $h$ when the ball was dropped.

Using the equation
s = ut +1/2at²
 h = 0 + 1/2 × 1.2 × (8.0)^2 = 38.4 m
Using $v = u + a t$
${u}_{b} = 1.2 \times 8.0 = 9.6 m {s}^{-} 1$

• Ball dropped from the elevator and simultaneously arrow shot from the ground.
Arrow.

Suppose the arrow hits the ball after $t$ from the time it was shot. It is also the same time after the ball was dropped. Let us assume that the arrow has traveled distance ${h}_{a}$ during this time. For sake of simplicity height of both persons has been ignored.

Given
${u}_{a} = 32 m . {s}^{-} 1$
a_a = –9.8 m.s^-2
${t}_{a} = t$

Using s = ut + 1/2at²
h_a = 32 t +1/2 xx (-9.8) × t^2 = 32 t – 4.9 t^2
h_a = 32 t – 4.9 t^2  ......(1)

Ball.
Given and calculated for the ball.
${u}_{b} = 9.6 m {s}^{-} 1$, (the instant of time when the ball was dropped it was traveling at the speed of the elevator).
$a = - 3.8 m {s}^{-} 2$, (the property"^$ of Styrofoam has been taken care of by this number, else the ball would have $a = - g$, the acceleration due to gravity). Using $s = u t + \frac{1}{2} a {t}^{2}$h_b = 9.6 × t +1/2 xx(- 3.8) × t^2 ${h}_{b} = 9.6 t - 1.9 {t}^{2}$.....(2) To make an assessment when and where does the arrow hit the ball. (a) Time ${t}_{1}$taken by the ball to reach maximum height from the point of drop Using $v = u + a t$$0 = 9.6 - 3.8 {t}_{1}$or ${t}_{1} = \frac{9.6}{3.8} = 2.53 s$Height ${h}_{1}$from the drop point, using ${v}^{2} - {u}^{2} = 2 a s$${0}^{2} - {9.6}^{2} = 2 \times \left(- 3.8\right) {h}_{1}$${h}_{1} = 12.13 m$Total height from the ground of ball at this point $=$Height at the point of drop $+$additional height gained $38.4 + 12.13 = 50.53 m$Distance traveled by arrow during this period. put $t = {t}_{1}$in (1) h_a = 32 t – 4.9 t^2  h_a = 32 xx2.53 – 4.9 xx2.53^2=49.6m  (b) It is clear that the arrow hits the ball only when it has started its downward journey from the position of highest point. Let the arrow hit the ball after elapse of time ${t}_{2}$from the point of maximum height attained by the ball. Let the height be ${h}_{m}$This can be found from (1) as h_m = 32 xx(2.53+t_2) – 4.9 (2.53+t_2)^2  .....(3) The ball moves down in this duration to meet the arrow$= 50.53 - {h}_{m}$Using $s = u t + \frac{1}{2} a {t}^{2}$$50.53 - {h}_{m} = 0 \times {t}_{2} + \frac{1}{2} \left(3.8\right) {t}_{2}$, we have ignored the sign convention and taken acceleration in the direction of motion. ${h}_{m} = 50.53 - 1.9 {t}_{2}$Inserting value of ${h}_{m}$in (3) 50.53-1.9t_2 = 32 xx(2.53+t_2) – 4.9 (2.53+t_2)^2 , solving for ${t}_{2}$ 80.96+32t_2 – 31.36-24.79t_2-4.9t_2^2 -50.53+1.9t_2=0 Ignoring ${t}_{2}^{2}$term as it is a small term from the geometry. Distance between the two is about $1 m$$9.11 {t}_{2} \approx 0.93$${t}_{2} \approx 0.10$$t = 2.53 + 0.10 \approx 2.63 s$"^$ Justification for taking given value of acceleration of the Styrofoam ball was already provided. More clarification:

When the ball is dropped. Three main forces come into play.
1. Gravity, downwards
2. Buoyancy, upwards
3. Drag, initially downwards; from the point of drop to the point when ball reaches maximum height. Thereafter upwards when the ball starts descent. Always opposite to the direction of velocity.

Per very fine analysis recently shared by fellow contributor Daniel W., contribution due to the buoyancy of Styrofoam in air is negligible as the density of Styrofoam varies from $10 \text{ to } 50 k g {m}^{-} 3$ against that $1.3 k g {m}^{-} 3$ of air. In light of this changes are warranted.

The statement of the question is silent about the drag. Keeping in with this drag has been treated as ignored.
We also know that
Drag$\propto - \text{velocity}$ for small particles and smaller velocities. and

Drag$\propto - {\text{velocity}}^{2}$ for larger velocities and larger objects.

In the instant case, keeping in view, the constant of proportionality, density of air, area of cross-section of the ball, decreasing magnitude of velocity upwards and very low value of velocity when the arrow hits the ball when it is descends could make a good case for ignoring Drag in comparison to Gravity.

There appears no real life justification for choosing such a low value of acceleration of the ball after dropping from the elevator.