Question

Two tall buildings are $40 \, m$ apart. With what speed must a ball be thrown horizontally from a window $145 \, m$ above the ground in one building, so that it will enter a window $22.5 \, m$ above from the ground in the other?

• A. $5 \, m \, s^{- 1}$
• B. $8 \, m \, s^{- 1}$
• C. $10 \, m \, s^{- 1}$
• D. $16 \, m \, s^{- 1}$

Answer 1

Answer: (B)

Given, $u_{x}=v$ and $u_{y}=0$

Separating the motion in x and y directions
Using $x=u_{x}t+\frac{1}{2}at^{2}$
$40=vt$ (i)
$\Longrightarrow t=\frac{40}{v}$ (ii)
$y=h_{\text{net}}=\text{145}-\text{22.5}$
Using $y=u_{y}t+\frac{1}{2}at^{2}$
$122.5=0+\frac{1}{2}\times 10t^{2}$
$122.5=5\left(\frac{40}{v}\right)^{2}$
$v=40\sqrt{\frac{5}{122.5}}=8 \, ms^{- 1}$