Question

You are walking around your neighborhood and you see a child on top of a roof of a building kick a soccer ball. The soccer ball is kicked at 51° from the edge of the building with an initial velocity of 13 m/s and lands 63 meters away from the wall. How tall, in meters, is the building that the child is standing on?

Answer 1

Answer:

$y_{o}=213m$

Explanation:

From the exercise we know that the child hits the ball with an initial velocity, its direction and where it hits the ground.

First of all, we need to calculate how long does it take to hit the ground

$x=v_{ox}t$

knowing that x=63m

$63m=(13cos(51)m/s)t$

$t=\frac{63m}{13cos(51)m/s}=7.70s$

Now, from free falling object's formula we know that position is:

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

When the ball hits the ground it is at y=0m

$0=y_{o}+(13sin(51)m/s)(7.70s)-\frac{1}{2}(9.8m/s)(7.70s)$

$y_{o}=(4.9m/s^2)(7.70s)^2-(13sin(51)m/s)(7.70s)=213m$

So, the building is 213m tall