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Question
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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Solution
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/s2
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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