# 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

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"#

#vecv_i = "7.3 m/s"#

#veca_x = 0#

#vecg = -"9.8 m/s"^2#

#Deltavecy = ?#

#Deltavecx = ?#

#vecv_f = ?#

**PART ONE: THE ASCENSION**

What I would do is find where the **apex** is to determine **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

#\mathbf(vecv_(fy)^2 = vecv_(iy)^2 + 2vecgDeltavecy)# where we say

#vecv_(fy) = 0# at the apex.

Since

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 part1.

Recall that a vertical velocity has a

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

Now that we have **free fall** is occurring.

The **total height** of the fall is **2**.

I get

**PART TWO: THE FREE FALL**

We can again treat the

At the apex, recall that **2**, and was the final velocity in part **1**. Now we can use another 2D kinematics equation. Remember that the total height is not

#\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

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

And like before, use a trig relation to get the

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

#t_"overall"# is NOT what we got in part2, 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

#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

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

Using

**PART FOUR: SOLVING FOR THE FINAL VELOCITY**

Now this is going to require a bit more thinking. We know that

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

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

Notice how we used **total height**, and we will take the *magnitude* of the total height for this.

And finally, since

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

where **1**. Now we just need to know what **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

Okay, we're ALMOST there. We are asked for **Pythagorean Theorem**.

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

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

Overall,

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

Here the vel. of projection,

the angle. of projection ,

The upward vertical component of vel of projection ,

The building being 17m high , the net vertical displacement reaching the ground will be

If the time of flight i.e.time for reaching ground is taken to be T

then using the formula

dividing both sides by 4.9 we get

(negative time discarded)

So Hero's Horizontal displacement before reaching ground will be

**Calculation of velocity at the time of reaching ground**

Vertical component velocity at the time of reaching ground

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

So resultant velocity at the time of reaching ground

Direction of

Is it helpful ?