in a volleyball game a player on one team spikes the ball over the net when the ball is 10 ft above the court. The spike drives the ball with a downward velocity of 55 feet per second. Players on the opposite team must hit the ball back over the net before the ball touches the court. How much time do the opposite players have to hit the spiked ball.

t = 3.61 s

Step-by-step explanation:

First, let's gather all the data. We know that the initial height (yi) is 10 ft. The final height (y) would be 0, Now, the speed (V) is 55 ft/s

The expression to use, in this kind of exercise (or the model to follow) is:

y = - 16t^2 + V*t + yi

You may wondering why we need to put V*t, this is because when you do the math and calculations, that expression becomes a distance.

Now, we have the speed, replacing data we have:

0 = - 16t^2 + 55t + 10 (1)

This is a second grade, and we need to use the general formula to solve this which is:

t = - b ± √b^2 - 4ac / 2a

Where a, b and c are the coefficients of the expression (1), in this case, 16, 55 and 10.

Replacing data here we have:

t = - 55 ± √ (55) ^2 - 4 * (-16) * (10) / 2 * (-16)

t = 55 ± √3025 + 640 / 32

t = 55 ± √3665 / 32

t = 55 ± 60.54 / 32 (2)

Now that we have (2), let's do the final math

t1 = 55 + 60.54 / 32 = 3.61 s

t2 = 55 - 60.54 / 32 = - 0.17 s

As we can't use negative time, we should take the first value, and the correct answer will be 3.61 s.