A volleyball is hit upward by a player in a game. The height h (in feet) is the volleyball after t seconds is given by the function h=-16t^2+30t+6.

Question

After how many seconds does the volleyball reach it’s maximum height? What is the maximum height of the volleyball? After how many seconds does the volleyball hit the ground?

Answer 1

If you haven't taken calculus yet you may not be familiar with finding or using derivatives. You can find the time when the ball reaches its maximum height by using the vertex formula:

x coordinate of vertex of a parabola = -^b/_2a

time of maximum height = -³⁰/_2*-16 = 0.9375

Also, the maximum height is not 48 feet, it's 20.0625 feet. The first term is negative due to the effects of gravity, so it's subtracted to give you 20.0625 feet

Answer 2

The ball will hit the ground when the height is zero.

0 = -16t² + 30t + 6

t = 2.05728 seconds or t = -0.18228 seconds

In this case, we want time to be positive since the ball will be hitting the ground after the player hits it in the air at time t = 0.

Therefore, t = 2.05728 seconds when the ball hits the ground.

The ball will hit its maximum height when the derivative is equal to zero.

h' = -16 * 2 * t + 30

h' = -32t + 30

0 = -32t + 30

-30 = -32t

30 = 32t

t = 0.9375 seconds when the ball is at its maximum height.

The height of the ball at the maximum height will be h(0.9375). The height equation with t = 0.9375 since that is the time where the ball is at its maximum.

h = -16(0.9375)² + 30(0.9375) + 6

h = 48.1875 feet at its maximum height.

Answer 3

You can solve this multiple ways. Solving by graphing is the easiest. You can open the desmos site for graphing. Type the equation in terms of x and not t. Where ever the max hump will be, this will be your maximum height. The x will be the seconds in the y will be the feet. Wherever the graph touches the x axis will be your seconds in which it hit the ground it will be (x,0).