A ball is thrown up and caught by a thrower after 4 seconds.how far below its highest point was the ball in 3 seconds after start?

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Vikhyat · Answer

Ball takes 4 sec to comes down so it mist be thrown with the velocity of 20m/s as g = 10m/s so it will take 2 sec to reach its highest point so applying second equation of motion s=ut+1/2at.^2 + 1 sec of coming down and will be 35 m

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A ball is thrown up and caught by a thrower after 4 seconds.how far below its highest point was the ball in 3 seconds after start?

A ball is thrown up and caught by a thrower after 4 seconds.how far below its highest point was the ball in 3 seconds after start?

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A ball is thrown up and caught by a thrower after 4 seconds.how far below its highest point was the ball in 3 seconds after start?

A ball is thrown up and caught by a thrower after 4 seconds.how far below its highest point was the ball in 3 seconds after start?

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