Question

A steel ball is dropped from a building’s roof and passes a window, taking 0.125 s to fall from the top to the bottom of the window, a distance of 1.20 m. It then falls to a sidewalk and bounces back past the window, moving from bottom to top in 0.125 s. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is 2.00 s. How tall is the building?

Answer 1

20.425 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

`s=ut+(1)/(2)at^2\\\Rightarrow u=(s-(1)/(2)at^2)/(t)\\\Rightarrow u=(1.2-(1)/(2)* 9.81* 0.125^2)/(0.125)\\\Rightarrow u=8.98\ m/s`

Now this will the final velocity when it reaches the top of the window

`v^2-u^2=2as\\\Rightarrow s=(v^2-u^2)/(2a)\\\Rightarrow s=(8.98^2-0^2)/(2* 9.81)\\\Rightarrow s=4.11\ m`

`v=u+at\\\Rightarrow v=8.98+9.81* 0.125\\\Rightarrow v=10.21\ m/s`

If the upward flight is an exact reverse of the downward flight then the time taken while going down will be 1 second and going up will be 1 second.

`s=ut+(1)/(2)at^2\\\Rightarrow s=10.21* 1+(1)/(2)* 9.81* 1^2\\\Rightarrow s=15.115\ m`

Hence total height of the building is 4.11+1.2+15.115 = 20.425 m