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प्रश्न
A ball is projected vertically upward with a speed of 50 m/s. Find the speed at half the maximum height. Take g = 10 m/s2.
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उत्तर
Given:
Initial speed of the ball, u = 50 m/s
Acceleration, a = −10 m/s2
At the highest point, velocity v of the ball is 0.
From v2 − u2 = 2 as, we have:
\[v = \sqrt{\left( u^2 + 2 as \right)}\]
\[ \Rightarrow v = \sqrt{\left( 50 \right)^2 + 2 \times \left( - 10 \right) \times \left( 62 . 5 \right)} = \sqrt{\left( 2500 - 1250 \right)}\]
\[ \Rightarrow v = \sqrt{1250} \approx 35 \text{ m } /s\]
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