Question

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

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

Answer 1

Answer: (B)

Given, $u_x=v$ and $u_y=0$

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