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A ball is thrown at a speed of 40 m/s at an angle of 60° with the horizontal. Find the maximum height reached  .

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प्रश्न

A ball is thrown at a speed of 40 m/s at an angle of 60° with the horizontal. Find the maximum height reached  .

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उत्तर

Given:
Initial speed of the ball, u = 40 m/s
Angle of projection of the ball with the horizontal, α = 60°
Also,
a = g = 10 m/s

Maximum height reached by the ball: 

\[H = \frac{u^2 \sin^2 \alpha}{2g}\]

\[\Rightarrow H = \frac{{40}^2 \left( \sin 60^\circ \right)^2}{2 \times 10} = 60 \text{ m } \]

 

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पाठ 3: Rest and Motion: Kinematics - Exercise [पृष्ठ ५२]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 3 Rest and Motion: Kinematics
Exercise | Q 33.1 | पृष्ठ ५२

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