Robert loves to dive . he joined the diving team . the diving board is feet above the water . if he jumps off the diving board as high as he can ( 10ft/s ) .how high above the water does he get ? and how long is he in mid air before he hits the water ?

It would help if you proofread your questions before you posted them.

How high is the diving board?

You have a typo and don't state how high above the water the diving board is.

I will give it an arbitrary value of 6 ft.
So the equation of height is

h = -16t^2 + 10t + 6, where h is in ft and t is in seconds , where t ≥ 0

for a max height, we need the vertex of that parabola
the t of the vertex is -b/(2a) = -10/-32 = .3125 seconds
and the height is -16(.3125^2) + 10(.3125) + 6
= 7.5625 ft

for how long in the air,
You want h = 0
-16t^2 + 10t + 6 = 0
divide by -2
8t^2 - 5t - 3 = 0
Using the formula
t = (5 ± √(25 - 4(8)(-3))/16
= (5 ± √121)/16
= -1/2 or 1

He will be in the air for 1 second

Adjust the above solution to whatever the height of the board is above the water.
I picked the 6 since I knew it would work out nice at the end.

To find out how high above the water Robert gets and how long he is in mid-air before hitting the water, we can use the kinematic equations of motion. Let's break down the problem step by step:

1. Find the maximum height Robert reaches:
The initial velocity, u = 10 ft/s (given)
The final velocity at the highest point, v = 0 ft/s (at the top of the trajectory)
Acceleration due to gravity, a = -32 ft/s² (negative because it opposes upward motion)
We can use the equation: v² = u² + 2a(s - u), where s is the distance traveled vertically.

Setting the equation up:
0² = 10² + 2(-32)(s - 0)
0 = 100 - 64s
64s = 100
s = 100/64
s ≈ 1.563 ft

Therefore, Robert reaches a maximum height of approximately 1.563 feet above the water.

2. Determine the time Robert is in mid-air:
To find the time taken to reach the highest point, we can use the equation: v = u + at.

Setting up the equation for the upward journey:
0 = 10 + (-32)t
-10 = -32t
t = -10/-32
t ≈ 0.3125 s

However, since we need to find the total time in mid-air (upward and downward journey), we double the calculated time:
Total time in mid-air = 2 * 0.3125 s
Total time in mid-air ≈ 0.625 s

Therefore, Robert is in mid-air for approximately 0.625 seconds before hitting the water.