## A girls throws a tennis ball straight into the air with a velocity of 64 feet/sec. If acceleration due to gravity is −32 ft/sec2, how many s

Question

A girls throws a tennis ball straight into the air with a velocity of 64 feet/sec. If acceleration due to gravity is −32 ft/sec2, how many seconds after it leaves the girl’s hand will it take the ball to reach its highest point? Assume the position at time t = 0 is 0 feet.

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6 months 2021-07-24T02:38:59+00:00 2 Answers 9 views 0

2 secs

Explanation:

Step 1:

Data obtained from the question:

Initial Velocity (U) = 64 feet/sec

Final velocity (V) = 0

acceleration due to gravity (g) = −32

ft/sec2

Time (t) =?

Step 2:

Determination of the time taken to reach the highest point. This is illustrated below:

V = U + gt

Applying the above equation, we can easily obtain the time (t) as follow:

0 = 64 + ( −32 x t )

0 = 64 – 32t

Collect like terms

0 – 64 = – 32t

– 64 = – 32t

Divide both side by -32

t = – 64/-32

t = 2 secs

Therefore, the time taken to reach the highest point is 2 secs

2 secs

Explanation:

Parameters given:

Initial velocity, u = 64 ft/s

Acceleration due to gravity, g = -32 ft/s (The minus sign indicates that the tennis ball is moving in the opposite direction of the gravitational force)

Using one of the equations of motion, we can find the time taken:

v = u + gt

When the ball reaches its highest point, the velocity (final velocity) is equal to zero, v = 0 ft/s

Therefore:

0 = 64 – 32*t

32t = 64

t = 64/32 = 2 secs

It takes two seconds for the ball to reach its highest point.