Question

A volleyball player hits a ball from a height of 6 feet with an initial vertical velocity of 32 feet per second. What is the maximum height of the volleyball? Assuming it is not hit by another player, when will the volleyball hit the ground? Round the answers to two decimal places if necessary.

The maximum height of the volleyball is ___ feet.

The volleyball will hit the ground after ___ seconds.

Answer 1

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This sounds like a projectile motion question. 

Horizontal velocity is always a constant (ignoring air resistance) so it shouldn't affect our answer. 

I'll assume vertical acceleration is -9.8 , for gravity

To find the maximum height of the volleyball, this is the split second when the ball stops travelling upwards and is suspended in the air before gravity brings it down again, i.e. when the vertical velocity is 0.

now we know initial vertical velocity = 32

so we can use v = v0-gt

where v = 0 and v0 = 32

0 = 32-9.8t

t= 3.27s

now we know how long it takes to get to the highest point, we want the height, or displacement.

y = 1/2*gt^2 + vt

y = 52.24

To find the maximum height, you add that amount to the original 6 feet to get 58.24 :)

To find when it hits the ground we need y=0

We can sub that into the formula, adding 6 to the end to include the original height

0 = 1/2(-9.8)t^2+32t +6

Using the quadratic formula to solve this quadratic

t = [-32 +/- √(32^2-4(-5.9)(6)]/-9.8

= 6.71 seconds because the answer has to be a positive

Hopefully all that’s correct :) I’m a bit rusty since it’s been a while since I’ve done projectile motion so let me know if I’ve made an error and I’ll be happy to review it again. 

Answer 2

Answer:

22 ft

2.27 seconds

Step-by-step explanation:

t = Time taken

u = Initial velocity = 32 ft/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32 ft/s²

$v=u+at\\\Rightarrow 0=32-32.2\times t\\\Rightarrow \frac{-32}{-32}=t\\\Rightarrow t=1 \s$

Time taken by the ball to reach the highest point is 1 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=32\times 1+\frac{1}{2}\times -32\times 1^2\\\Rightarrow s=16\ ft$

The highest point reached by the ball above its release point is 16 ft

The maximum height from the ground is 6+16 = 22 feet

$s=ut+\frac{1}{2}at^2\\\Rightarrow 22=0t+\frac{1}{2}\times 32\times t^2\\\Rightarrow t=\sqrt{\frac{22\times 2}{32}}\\\Rightarrow t=1.17\ s$

Total time taken for the ball to hit the ground after release is 1+1.17 = 2.27 seconds