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 $2 \,m$ 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 $t_{1}=2 t_{2}$, 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 $v_{1}$
$h_{1}=\frac{v_{1}^{2}}{2 g}$
Another body of mass $2\, m$ is projected with a velocity $v_{2}$ at an angle $\theta$.
$\therefore $ Height attained $\left(h_{2}\right)=\frac{v_{2}^{2} \sin ^{2} \theta}{2 g}$
$\therefore \frac{h_{1}}{h_{2}}=\frac{v_{1}^{2}}{v_{2}^{2}} \sin ^{2} \theta$
But $t_{1}=2 t_{2}$
$\therefore \frac{v_{2}}{g}=2\left[\frac{v_{2} \sin \theta}{g}\right]$
$\Rightarrow v_{1}=2 v_{2} \sin \theta$
$\therefore \frac{h_{1}}{h_{2}}=\left(\frac{2}{1}\right)^{2}=\frac{4}{1}$