A soccer ball is kicked from 3 feet above the ground. The height, h, in feet of the ball after t seconds is given by the function. h=-16t^2+64t+3.

Question

Find the time it takes for the ball to return back to ground level.

 

Find the maximum height reached by the ball.

Answer 1

(a) This is a parabolic function where the parabola opens downward showing that the ball will rise initially to a maximum height then come back down.  The ball is at ground level where -16t² + 64t + 3 = 0

 

Lets use the quadratic equation where given ax² + bx + c + 0 is given by

x = -b/2a +- (sqrt(b²-4ac))/2a

a = -16, b = 64, c = 3

 

t = -64/(2(-16)) +- (sqrt(64² - 4(-16)(3))/(2(-16))

t = -64/(-32) +- (sqrt(4096+192))/(-32)

t = 2 +- (sqrt(4288))/-32

t = 2 +- 65.483/(-32)

t = 2 +- 2.046

 

so this parabola would hit zero at times t = 4.046sec and -0.046 sec

We can discard the t = -0.046 because that arc of the parabola is actually to the left of our starting point and is not applicable.

Thus the ball would rise for a couple seconds, then start descending, and be at ground level when t = 4.046sec

 

(b)  The maximum height of the ball would be reached at the vertex which is where t = -b/2a which is at t = 2 secs

 

Plugging t = 2 into the original equation we get h = -16(2)² + 64(2) + 3

h = -16(4) + 128 + 3

h = -64 + 128 + 3

h = 67

 

the maximum height reached by the soccer ball is 67ft at time t = 2 seconds