Question

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?

Answer 1

Answer:

the ball left the boy's hand with 39.2 m/sec speed

Answer 2

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 $9.8 m/s^2$

At maximum height the final velocity of the ball, v =0.

Substitute the given values in above equation we get,

$0 = u -9.8m/s^2(4 s)$

$u = 39.4 m/s$

Thus, the velocity of the ball when it left the boy`s hand is 39.4 m/s.

Learn more : first equation of motion,

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