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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 range of the ball. Take g = 10 m/s2.
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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
Horizontal range of the ball:
\[R = \frac{u^2 \sin2\alpha}{g}\]
\[= \frac{{40}^2 \sin\left( 2 \times 60^\circ\right)}{10} = 80\sqrt{3} \text{ m } \]
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