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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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प्रश्न

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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उत्तर

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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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 32.3 | पृष्ठ ५२

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