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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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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 of the ball is 0.

\[s = \frac{125}{2} = 62 . 5 \text{ m } \]

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

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 3 Rest and Motion: Kinematics
Exercise | Q 23.3 | पृष्ठ ५२

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