Question

9) Joshua threw a ball that traveled on a parabolic path and then landed on the ground. Its flight is modeled

by the equation, h(t) = -5t² + 20t+ 0.4, where time, t, is in seconds and height, h, is in meters.
a) Determine the maximum height reached by the ball.
b) How long did it take to reach maximum height?
c) When will the ball hit the ground?

Answer 1

a) The maximum height reached by the ball can be found by taking the derivative of the given equation, setting it equal to 0, and solving for t. We have:

h'(t) = -10t + 20 = 0

Solving for t, we get t = 2. Substituting this value into the original equation, we get:

h(2) = -5(2)² + 20(2) + 0.4 = -10 + 40 + 0.4 = 30.4

Thus, the maximum height reached by the ball is 30.4 meters.

b) The time it takes to reach the maximum height can be found by setting the derivative of the given equation equal to 0 and solving for t, as we did in part (a). In this case, we get t = 2, so it takes 2 seconds for the ball to reach its maximum height.

c) The time at which the ball hits the ground can be found by setting the original equation equal to 0 and solving for t. We have:

h(t) = -5t² + 20t + 0.4 = 0

Solving for t, we get:

-5t² + 20t + 0.4 = 0

This quadratic equation has two solutions, which can be found using the quadratic formula:

t = (-20 ± √(20² - 4(-5)(0.4))) / (2(-5))

= (-20 ± √(400 + 0.8)) / -10

= (20 ± √400.8) / -10

= (-20 ± 20.28) / -10

= -10 ± 10.14

Thus, the ball hits the ground at t = -10 + 10.14 = 0.14 seconds, or at t = -10 - 10.14 = -20.14 seconds. Since the time must be positive, we can conclude that the ball hits the ground at 0.14 seconds.