A bowling machine placed at a height $h$ above the earth surface releases different balls with different…
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Solution

$\begin{aligned} & u_x=v_x=10 \sqrt{3} m s^{-1} \\ & t=\sqrt{\frac{2 h}{g}} \\ & \therefore v_y=u_y+g t=0+g \sqrt{\frac{2 h}{g}} \\ & \therefore \tan 30^{\circ}=\frac{v_y}{v_x}=\frac{g \sqrt{\frac{2 h}{g}}}{10 \sqrt{3}}=\frac{\sqrt{2 g h}}{10 \sqrt{3}} \\ & \therefore \quad h=5 m\end{aligned}$
Asked in: AP EAMCET 2024 (23 May Shift 1)
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