Question

A skateboarder is trying to skate horizontally off a building and jump over a pool as shown! The pool is 25.0m wide and the roof of the building is 32.0m above where he will land. Neglect drag.
How long will the skateboarder remain in the air?

With what initial speed (in given units) does the skateboarder need to barely clear the pool?

Would the skateboarder clear the pool if he skated off the building with an initial speed of 27mph?

Answer 1

Answer:

a) t = 2.55s

b) $v_{0x} = 9.80 m/s$

c) yes

Explanation:

In order to solve this problem, we can start by drawing a sketch of the situation so we can better visualize what the problem is about (see attached picture).

a)

For the first question. We are talking about a movement in two dimensions. So on the first question they are asking us for vertical movement. It will be uniformly accelerated, so we can use the following formula:

$y_{f}=y_{0}+V_{0y}t+\frac{1}{2}at^{2}$

We know the following:

$y_{f}=0$

$y_{0} = 32m$

$V_{0y}=0$

t=?

$a=-9.81 m/s ^{2}$

With this data, we can simplify our equation, so we end up with:

$y_{0}+\frac{1}{2}at^{2}=0$

so we can now substitute the data we know and solve for t:

$32m-\frac{1}{2}(9.81 m/s^{2})t^{2}=0$

$(-4.905 m/s^{2})t^{2}=-32m$

$t^{2} = \frac{-32m}{-4.905 m/s^{2}}$

$t=\sqrt{6.52s^{2}}$

t = 2.55 s

b)

For part b, since we are talking about horizontal movement and we are neglecting drag, this means that his horizontal speed will be constant. So we can use the following formula:

$V_{x}=\frac{x}{t}$

we know he most move a horizontal distance of 25 meters in a time of 2.55s so we get:

$V_{x} = \frac{25m}{2.55s}$

$V_{x}=9.80 m/s$

c) for part c, we can do the conversion between miles per hour to meters per second like this:

$\frac{27mi}{1hr}*\frac{1hr}{3600s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}$

so the given initial speed is equivalent to:

12.07 m/s

this is greater than the minimum 9.80 m/s we need, so the skater will clear the pool at this speed.