A ball rolls horizontally off the edge of a tabletop that is 2.00 m high. It strikes the floor at a point 1.69 m horizontally away from the table edge. (Neglect air resistance.) (a) How long was the ball in the air? S (b) What was its speed at the instant it left the table? m/s

Question

29-04-2025
Physics

High School

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A ball rolls horizontally off the edge of a tabletop that is 2.00 m high. It strikes the floor at a point 1.69 m horizontally away from the table edge. (Neglect air resistance.) (a) How long was the ball in the air? S (b) What was its speed at the instant it left the table? m/s

Answer :

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kumargaurav1012000 kumargaurav1012000 · Answer 1 · 2024-04-29T17:48:36-04:00

a) To calculate the time taken by the ball in the air, we can use the formula for vertical displacement, S_y. Since the initial vertical velocity, u_y, is zero when the ball is thrown horizontally off the table, we can simplify the equation to:

S_y = 1/2 * a_y * t^2

Where S_y is the height of the table (2.00 m), a_y is the acceleration due to gravity (-9.81 m/s^2), and t is the time taken by the ball to reach the ground level.

Plugging in the values, we have:

2.00 = 1/2 * (-9.81) * t^2

Solving for t, we find t = 0.638 s.

Therefore, the time taken by the ball in the air is approximately 0.638 s.

b) To calculate the speed of the ball when it left the table, we can use the formula for horizontal displacement, S_x, and the time taken, t. Given that S_x is 1.69 m and t is 0.638 s, we can find the initial horizontal component of velocity, u_x:

u_x = S_x / t = 1.69 / 0.638 = 2.65 m/s

Hence, the speed of the ball when it left the table was approximately 2.65 m/s.

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