# An object is thrown vertically from a height of 2 m at 7 m/s. How long will it take for the object to hit the ground?

Mar 12, 2018

It should take 1.6726 seconds.

#### Explanation:

The problem needs to be broken down into two parts: the time it takes the object to reach the apex of the throw (where v=0) and the time it takes to hit the ground from the apex.

Part 1: Time to apex
Assuming a gravitational constant of $g = - 9.8 \frac{m}{{s}^{2}}$ we can use classical Newtonian physics to get the time and distance traveled. We know that at the apex v=0:

${V}_{f} = {V}_{i} + a t$
${V}_{f}^{2} = {V}_{i}^{2} + 2 a d$

$0 = 7 - 9.8 t \Rightarrow t = 0.7143 s$
$0 = {7}^{2} - 2 \cdot 9.8 \cdot d \Rightarrow d = 2.5 m$

We need to know d for the second part, since the object will have to fall d+2 meters (since the object started 2m up)

For the second part, the object will fall 4.5m, with an initial velocity of 0. Since we're traveling in the same direction as gravity, the constant is now $+ 9.8 \frac{m}{s} ^ 2$:

${V}_{f}^{2} = {V}_{i}^{2} + 2 a d$
${V}_{f}^{2} = 0 + 2 \cdot \left(9.8\right) \left(4.5\right) \Rightarrow {V}_{f} = \sqrt{88.2} = 9.3915 \frac{m}{s}$

${V}_{f} = {V}_{i} + a t$
$9.3915 = 0 + 9.8 t \Rightarrow t = 0.9583 s$

Finally we add the two times together to get the total time:

$0.7143 + 0.9583 = 1.6726 s$

Mar 12, 2018

$1.66 s$

#### Explanation:

Apply the equation,$s = u t + \frac{1}{2} a {t}^{2}$ for solving this problem,when the object reaches the ground it's displacement will be $- 2 m$,(considering downward direction to be negative),as it first goes up and then comes back to the point of projection,so displacement is only for going down from the point of projection.

So,if it takes time $t$ to reach the ground we can write,

-2=7t -1/2×9.8t^2

Solving it we get, $t = 1.66 s$

Mar 12, 2018

If thrown vertically up, $\text{T = 1.64 s}$

If thrown vertically down, $\text{T = 0.24 s}$

#### Explanation:

Object is thrown vertically up or vertically down?

Case-1: Object thrown vertically up

Time taken to reach the initial point will be

t_1 = (2u)/g = (2 × "7 m/s")/"9.8 m/s"^2 = "1.4 s"

Velocity when it hits the ground

$v = \sqrt{{u}^{2} + 2 a d}$

$v = \sqrt{\left(\text{7 m/s")^2 + (2 × "9.8 m/s"^2 × "2 m}\right)}$

$v = \text{9.4 m/s}$

Further time taken to reach the ground

t_2 = (v - u)/g = ("9.4 m/s "-" 7 m/s")/"9.8 m/s"^2 = "0.24" s

Total time taken $= {t}_{1} + {t}_{2} = \text{1.4 s + 0.24 s" = color(blue)"1.64 s}$

Case-2: Object thrown vertically down

Velocity when it hits the ground

$v = \sqrt{{u}^{2} + 2 a d}$

$v = \sqrt{\left(\text{7 m/s")^2 + (2 × "9.8 m/s"^2 × "2 m}\right)}$

$v = \text{9.4 m/s}$

Time taken to reach the ground

T = (v - u)/g = ("9.4 m/s "-" 7 m/s")/"9.8 m/s"^2 = color(blue)"0.24 s"