Question

A steel ball is dropped from a building's roof and passes a window, taking 1s to fall from the top to the window to the bottom of the window, a distance of 10m. It then falls to a sidewalk and bounces back past the window, moving from bottom to top in 1s. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is 2s. How tall is the building?

• A. 33.5 m
• B. 30 m
• C. 25.6 m
• D. 31.25 m

Answer 1

The correct option is D

31.25 m

BC is the window. When the ball falls from B to C.

$s=-10m,a=-10m/s^2,t=1s$

$\therefore s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow -10=u\left(1\right)+\frac{1}{2}(-10)(1{)}^2$

$\Rightarrow -10=u-5$

$\Rightarrow u=-5m/s$

$\therefore$Velocity at B, =-5m/s

Velocity at C = u + at = -5-10(1) = -15 m/s

Time for CD is 1s

$\therefore$ CD = s = $ut+\frac{1}{2}a{t}^2$

$-15\left(1\right)+\frac{1}{2}(-10)(1{)}^2$

=-15 - 5 = 20m $\therefore$CD = 20m

To find AB

$v^2=u^2+2as$

$\Rightarrow (-5{)}^2=0+2(-10\left)\right(s)$

$\Rightarrow 25=0-20s$

$\Rightarrow s=\frac{-25}{20}=\frac{-5}{4}=-1.25m\therefore AB=1.25m$

$\therefore$ TotaL height = 1.25 + 10 + 20 = 31.25m