A ball thrown horizontally at 22.0 m/s from the roof of a building lands 40.0 m from the base of the building. How high is the building?

Question

How do you solve this?

Answer 1

let's think air is static and not influencing the motion of the ball.

Thus the ball has an initial velocity v₀=22.0 m/s and vertical acceleration, g= acceleration due to gravity= 9.8 m/s 

It covers horizontal distance, x=40.0 m supposedly in t seconds

Since the ball has been thrown horizontally the motion path of the ball, the projectile, yields an angle of 0 degree with the horizontal axis of references.

The vertical distance, say, y, traveled in time t seconds hence the height of the building.

Hence we have, 40=22tcos0

=> 40= 22t ... since cos0=1

=>t=40/22

y= 22tsin0-1/2(gt²) m

 =0-0.5(9.8)(40/22)² m  ............ since sin0=0

 =-(4.9)(1.818)² m ............. rounded up to 3 decimal place

 =-(4.9)(3.305124) m

 =-16.1951076 m

 =-16.195 m

We can ignore the negative sign as it denotes the downward motion of the ball.

Therefore, rounding up to three decimal place, the building seems 16.195 meter high.