Question

A ball of mass m is thrown vertically upwards another ball of mass 2m is thrown at an angle theta with the vertical both of them stay in air for same period of time the heights attained by the two balls are in the ratio of

Answer 1

Hey

ATQ, let's take the velocities ( of both balls ) as $v_0 \ and \ v_1$

Since time taken by both is equal, from the first law of motion, we see Vertical component of both velocities are equal ... 

$v_1cos \theta = v_0 \\ \\ Now, we \ also \ know \ 2gs = v^2 - u^2 \\ \\ From \ this, we \ derive :-\ \textgreater \ \\ \\ s_0 = \frac{v_0^2}{2g} \\ \\ s_1 = \frac{v_1^2cos^2 \theta }{2g} \\ \\ Hence, the \ ratio \ is \ \ \ \ \frac{s_1}{s_0} = \frac{\frac{v_1^2cos^2 \theta }{2g}}{ \frac{v_0^2}{2g} } = \frac{v_1^2cos^2 \theta}{v_0^2}$

Hope it helps ^_^

Answer 2

Answer:

See the attachment below

Explanation: