Question

A person is throwing two balls into the air, one after the other. He throws the second ball when the first ball is at the highest point. If he is throwing the balls at an interval of 2 seconds, how high do they rise?
$[Takeg=10m/s^2]$

• A. 5 m
• B. 20 m
• C. 2.50 m
• D. 1.25 m

Answer 1

The correct option is B 20 m

The person throws the second ball when the first reaches its maximum height.
Hence, time of ascent of the first ball = 2s
We know, time of ascent is half of total time of flight. Then, we get
$\frac{T}{2}=\frac{u}{g}$
$2=\frac{u}{g}\Rightarrow u=20m/s$
Using this we can find the maximum height
$H=\frac{u^2}{2g}$
$H=\frac{20\times 20}{2\times 10}=20m$