A superhero launches himself from the top of a building with a velocity of 7.3m/s at an angle of 25 above the horizontal. If the building is 17 m high, how far will he travel horizontally before reaching the ground? What is his final velocity?

2 Answers
Feb 27, 2016

A diagram of this would look like this:

What I would do is list what I know. We will take negative as down and left as positive.

h = "17 m"h=17 m
vecv_i = "7.3 m/s"vi=7.3 m/s
veca_x = 0ax=0
vecg = -"9.8 m/s"^2g=9.8 m/s2

Deltavecy = ?
Deltavecx = ?
vecv_f = ?

PART ONE: THE ASCENSION

What I would do is find where the apex is to determine Deltavecy, and then work in a free fall scenario. Note that at the apex, vecv_f = 0 because the person changes direction by virtue of the predominance of gravity in decreasing the vertical component of the velocity through zero and into the negatives.

One equation involving vecv_i, vecv_f, and vecg is:

\mathbf(vecv_(fy)^2 = vecv_(iy)^2 + 2vecgDeltavecy)

where we say vecv_(fy) = 0 at the apex.

Since vecv_(fy)^2 < vecv_(iy)^2 and Deltavecy > 0, Deltavecv_y^2 < 0 and this equation is indeed asking us to use g < 0.

For part 1:

color(blue)(Deltavecy) = (vecv_(fy)^2 - v_(iy)^2)/(2g) = color(blue)((-v_(iy)^2)/(2g)) > 0

where vecv_(fy) = 0 is the final velocity for part 1.

Recall that a vertical velocity has a sintheta component (draw a right triangle and get the sintheta = (vecv_(y))/(vecv) relationship).

color(green)(Deltavecy = (-v_(i)^2 sin^2theta)/(2g)) > 0

Now that we have Deltavecy and we know that vecv_y has changed direction, we can suppose free fall is occurring.

The total height of the fall is color(green)(h + Deltavecy). That is something we can use for part 2.

I get Deltavecy to be about "0.485 m" and h + Deltavecy to be about color(blue)("17.485 m").

PART TWO: THE FREE FALL

We can again treat the y direction independently of the x direction, since veca_x = 0.

At the apex, recall that color(green)(vecv_(iy) = 0), which is the initial velocity for part 2, and was the final velocity in part 1. Now we can use another 2D kinematics equation. Remember that the total height is not Deltavecy here!

\mathbf(h + Deltavecy = 1/2g t_"freefall"^2) + cancel(v_(iy)t_"freefall")^(0)

Now we can just solve for the time it takes to hit the ground from the apex.

color(green)(t_"freefall") = sqrt((2(h + Deltavecy))/g)

= color(green)(sqrt((2(h - (v_(i)^2 sin^2theta)/(2g)))/g))

and of course, time is obviously not ever negative, so we can ignore the negative answer.

...And we're getting there.

PART THREE: SOLVING FOR THE HORIZONTAL DISTANCE

We can reuse the same kinematics equation as the one previously examined. One of the things we have been going for is Deltax, which is:

color(blue)(Deltax) = cancel(1/2a_xt^2)^(0) + v_(ix)t

And like before, use a trig relation to get the x component (costheta).

= color(blue)(vecv_icostheta*t_"overall") > 0

where t_"overall" is NOT what we got in part 2, but will include the time t_"leap" going from the building to the apex of the flight and t_"freefall" that we acquired earlier.

Deltay = 1/2vecg t_"leap"^2 + vecv_(iy)t_"leap"

With Deltay ~~ "0.485 m". When we solve this using the quadratic equation, it would yield:

t_"leap" = (-(vecv_(iy)) + sqrt((vecv_(iy))^2 - 4(1/2vecg)(-|Deltay|)))/(2*1/2vecg)

~~ "0.3145 s"

Include the time acquired for apex to the ground and you should get about color(blue)("2.20 s") for the entire flight. Let's call this t_"overall".

t_"overall" = t_"leap" + t_"freefall"

Using t_"overall", I get color(blue)(Deltavecx ~~ "14.58 m").

PART FOUR: SOLVING FOR THE FINAL VELOCITY

Now this is going to require a bit more thinking. We know that h = "17 m" and we have Deltax. Therefore, we can determine the angle with respect to the horizontal ground.

tantheta' = (h+Deltavecy)/(Deltavecx)

color(blue)(theta' = arctan((h+Deltavecy)/(Deltavecx)))

Notice how we used h + Deltavecy since we did in fact jump upwards before falling, and we did not jump straight forward. So, the angle theta involves Deltax and the total height, and we will take the magnitude of the total height for this.

And finally, since vecv_x has not changed all this time (we ignore air resistance here):

color(green)(vecv_(fx)) = vecv_(ix) = vecv_fcostheta' = color(green)(vecv_icostheta') > 0

where vecv_i is the initial velocity from part 1. Now we just need to know what vecv_(fy) is in part 2. Go back to the beginning to see:

vecv_(fy)^2 = cancel(vecv_(iy)^2)^(0) + 2vecg*(h+Deltavecy)

Hence, this becomes:

color(green)(vecv_(fy) = -sqrt(2vecg*(h+Deltavecy))) < 0

Remember that we defined down as negative, so h+Deltay < 0.

Okay, we're ALMOST there. We are asked for vecv_f. Therefore, we finish by using the Pythagorean Theorem.

vecv_f^2 = vecv_(fx)^2 + vecv_(fy)^2

color(blue)(vecv_f = -sqrt(vecv_(fx)^2 + vecv_(fy)^2)) < 0

Overall, color(blue)(|vecv_f| ~~ "19.66 m/s").

And that would be all of it! Check your answer and tell me if it worked out.

Feb 28, 2016

Here the vel. of projection, v=7.3ms^-1
the angle. of projection ,alpha=25^0 above horizontal
The upward vertical component of vel of projection ,vsinalpha= 7.3*sin25^0 =7.3*0.42ms^-1~~3.07ms^-1

The building being 17m high , the net vertical displacement reaching the ground will be h=-17m as the superhero projected himself upward (taken positive here)

If the time of flight i.e.time for reaching ground is taken to be T
then using the formula h = vsinalpha*t-1/2*g*t^2 we can have
=>-17=3.07*T-0.5*9.8*T^2
=>4.9T^2-3.07T-17=0

dividing both sides by 4.9 we get

=>T^2-0.63T-3.47=0
=>T=(0.63+sqrt((-0.63)^2-4*1*(-3.47)))/2~~2.20s
(negative time discarded)

So Hero's Horizontal displacement before reaching ground will be
=T*vcosalpha=2.20**7.3cos(25^0)~~14.56m

Calculation of velocity at the time of reaching ground

Vertical component velocity at the time of reaching ground

v_y^2= u^2sin^2alpha+2xx(-9.8)xx(-17)

Again horizontal component of the velocity at the time of reaching ground

=>v_x=ucosalpha

So resultant velocity at the time of reaching ground

v_r=sqrt(v_x^2+v_y^2)=sqrt(u^2sin^2alpha+u^2cos^2alpha-2xx9.8xx17)

=>v_r=sqrt(u^2+2xx9.8xx17)

=>v_r=sqrt(7.3^2+2xx9.8xx17)=19.66"m/s"

Direction of v_r with the horizontal=tan^-1(v_y/v_x)

=tan^-1(sqrt(u^2sin^2alpha+2xx(-9.8)xx(-17))/(ucosalpha))

=tan^-1(sqrt(7.3^2sin^2 25+2xx(-9.8)xx(-17))/(7.3cos25))
=70.3^@->"downward with the horizontal"

Is it helpful ?