Question

A ball of mass $m$ is thrown vertically upwards. Another ball of mass $2m$ is thrown up making an angle $\theta$ with the vertical. Both of them stay in air for the same time. Their maximum heights are in the ratio of

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

Answer 1

The correct option is B $1:1$

$T_1=\frac{2u_1}{g}$

$T_2=\frac{2u_2\sin \theta }{g}$

Given: $T_1=T_2$

$\Rightarrow \frac{2}{g}{u}_1=\frac{2u_2\sin \theta }{g}\Rightarrow u_1=u_2\sin \theta -\left(1\right)$

Height $=\frac{u^2{\sin }^2\theta }{2g}\Rightarrow \frac{H_1}{H_2}=\frac{u_1^2}{u_2^2{\sin }^2\theta }$

using (1) $\Rightarrow \frac{H_1}{H_2}=1$