In a volleyball game, a player on one team spikes the ball over the net when the ball is 10 feet above the court. the spike drives the ball downward with an initial vertical velocity of 55 feet per second. how much time does the opposing team have to return the ball before it touches the court? round your answer to the nearest hundredth.

4.75 seconds

Step-by-step explanation:

Initial height of the ball from the ground = 10 feet

Upward initial velocity given = 55 feet per second

Value of g ( acceleration due to gravity) = 32.2 feet per s²

Motion of the ball:

The ball first goes vertically upwards, gravity decelerates the body and it momentarily comes to rest at a point in its upward trajectory, which is the point of maximum height. From this point, to the ground, the ball behaves as a freely dropped body.

Till the ball reaches its maximum height:

u = + 55

v = 0 ( final velocity is zero)

a = - 32.2 (since it is deceleration)

we know that v = u + at

⇒    0 = 55 - 32.2t

⇒  t = 1.7 s

Also , we have s = ut + (1/2)at²

Here, s is the maximum height from the point where ball is thrown

So,     s = 55(1.7) - (0.5)(32.2)(1.7)(1.7)

⇒ s = 140 feet

So at a height of 140 feet + 10 feet (initial height) = 150 feet, the ball acts as a freely dropped body.

Here,      u = 0

               a = +32.2

               s= 150

s = ut + (1/2)at²

⇒  150 = 0.5 ( 32.2) (t²)

⇒  t² = 300/32.2 = 9.31

⇒ t = 3.05 sec

So total time = 1.7 + 3.05 = 4.75 seconds

3

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