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

A bowling machine placed at a height $h$ above the earth surface releases different balls with different angles but with same velocity $10 \sqrt{3} \mathrm{~ms}^{-1}$. All these balls landing velocities make angles $30^{\circ}$ or more with horizontal. Then the height ' h ' (in meters) (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
  1. 15
  2. 12
  3. 10
  4. 5

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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