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Question
A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m/s. Find the velocity (direction and magnitude) with which it strikes the ground.
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Solution
Given:
Speed of the ball, ux = 20 m/s
Height from which the ball is dropped, h = 100 m
We know that horizontal velocity remains constant throughout the motion of the ball.
At A, vx = 20 m/s.
vy = u + gt = 0 + 9.8 × 4.5
⇒vy = 44.1 m/s
Resultant velocity:
\[v_r = \sqrt{\left( 44 . 1 \right)^2 + \left( 20 \right)^2} \approx 49 \text{ m } /s\]
\[\tan \beta = \frac{v_y}{v_x} = \frac{44 . 1}{20} = 2 . 205\]
\[ \Rightarrow \beta = \tan^{- 1} \left( 2 . 205 \right) = 66^\circ \]
Therefore, the ball strikes the ground with a magnitude of velocity 49 m/s and the direction at an angle of 66° with the ground.
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