JEE MAIN - Physics (2022 - 27th July Morning Shift - No. 26)
A ball of mass m is thrown vertically upward. Another ball of mass $$2 \mathrm{~m}$$ is thrown at an angle $$\theta$$ with the vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is $$\frac{1}{x}$$. The value of x is _____________.
Answer
1
Explanation
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$$\therefore$$ $${u_1} = {u_2}\sin \theta $$
$${{{H_1}} \over {{H_2}}} = {{{{u_1^2} \over {2g}}} \over {u_2^2{{{{\sin }^2}\theta } \over {2g}}}}$$
$$ = {\left( {{{{u_1}} \over {{u_2}\sin \theta }}} \right)^2} = 1$$
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