A boy throws a ball straight up into the air. It reaches the highest point of its flight after 4 seconds. How fast was the ball going when it left the boy's hand?
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Answer:
the ball left the boy's hand with 39.2 m/sec speed
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The velocity of the ball when it left the boy`s hand is 39.4 m/s.
Explanation:
Given time taken by ball to reach the maximum height, t = 4 s.
To calculate the initial velocity of the ball when it left the boy`s hand, use first equation of motion,
v = u +gt
Here, v is initial velocity u is final velocity t is the time taken and g is acceleration due to gravity and its value is
At maximum height the final velocity of the ball, v =0.
Substitute the given values in above equation we get,
Thus, the velocity of the ball when it left the boy`s hand is 39.4 m/s.
Learn more : first equation of motion,
https://brainly.in/question/12883484
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