Question

A ball of mass $M$ is thrown vertically upwards. Another ball of mass $2 M$ is thrown at an angle $\theta$ with the vertical. Both of them stay in air for the same period of time. The heights attained by the two are in the ratio

• A. 1: 2
• B. 2: 1
• C. 1: 1
• D. $1: \cos \theta$

Answer 1

Answer: (C)

For the ball thrown vertically upwards, the time taken by the ball to come back is $T_{1}=\frac{2 u_{1}}{g}$.
For the ball projected at an angle $\theta$ with the vertical, the time of flight is $T_{2}=\frac{2 u_{2} \cos \theta}{g}$
Since time for both the balls is same, so
$\frac{2 u_{1}}{g}=\frac{2 u_{2} \cos \theta}{g} $ or
$ u_{1}=u_{2} \cos \theta $
Now, $ h_{1}=\frac{u_{1}^{2}}{2 g} $ and
$ h_{2}=\frac{u_{2}^{2}}{2 g} \cos ^{2} \theta$
Hence, $\frac{h_{1}}{h_{2}}=\frac{u_{1}^{2}}{u_{2}^{2} \cos ^{2} \theta}$
$=\frac{u_{2}^{2} \cos ^{2} \theta}{u_{2}^{2} \cos ^{2} \theta}=1 $ (Using (i))