Question

A ball of mass $m$ is thrown vertically upwards. Another ball of mass $2m$ is thrown simultaneously at an angle $\theta$ with horizontal. Both of them hit the ground at the same time when thrown from the same height. The heights attained by the two are in the ratio ?

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

Answer 1

Answer: (B)

1 : 1

Let $h_1$ be the maximum height reached by ball 1 and $h_2$ be the height reached by ball 2

$v_1^2=u_1^2-2g{h}_1(v_1=0)$

$h_1=\frac{u_1^2}{2g}$

$v_2^2=u_2^{2}si{n}^2\theta -2g{h}_2$

$h_2=\frac{u_2^{2}si{n}^2\theta }{2g}$

Time of accent for both the bodies is same. Therefore

$v_1=u_1-g{t}_1$

$t_1=\frac{u_1}{g}$

$t_2=\frac{u_{2}sin\theta }{g}$

$t_1=t_2$ (given in question)

$u_1=u_{2}sin\theta$

$\frac{h_1}{h_2}=\frac{\frac{u_1^2}{2g}}{\frac{u_2^{2}si{n}^2\theta }{2g}}$

Substituting $u_1=u_{2}sin\theta$

We get $h_1=h_2$

Thus, the heights attained by the two are in the ratio $1:1$.