Question

A body of mass $m$ thrown up vertically with velocity $v_1$ reaches a maximum height $h_1$ in $t_1$ seconds. Another body of mass $2m$ is projected with a velocity $v_2$ at an angle $\theta$. The second body reaches a maximum height $h_2$ in time $t_2$ seconds. If $t_1=2t_2$, then ratio $\left(\frac{h_1}{h_2}\right)$ is

• A. 1: 2
• B. 4: 1
• C. 1: 1
• D. 3: 2

Answer 1

Answer: (B)

Maximum vertical height attained by body thrown with velocity $v_1$
$h_1=\frac{v_1^2}{2g}$
Another body of mass $2m$ is projected with a velocity $v_2$ at an angle $\theta$.
$\therefore$ Height attained $\left(h_2\right)=\frac{v_2^2{\sin }^2\theta }{2g}$
$\therefore \frac{h_1}{h_2}=\frac{v_1^2}{v_2^2}{\sin }^2\theta$
But $t_1=2t_2$
$\therefore \frac{v_2}{g}=2\left[\frac{v_2\sin \theta }{g}\right]$
$\Rightarrow v_1=2v_2\sin \theta$
$\therefore \frac{h_1}{h_2}={\left(\frac{2}{1}\right)}^2=\frac{4}{1}$