Question

A ball is launched into the air from a height of 12 feet at time t = 0. The function that models this situation is h(t) = -2t2 + 10t + 12, where t is measured in seconds and h is the height in feet.
Part A (1 Point)
What is the height of the ball after 2 seconds?

Part B (3 Points)
When will the ball hit the ground?

Answer 1

Answer:

(a) 24 ft

(b) 2.5 s

Step-by-step explanation:

At t=2

h=-2(2)^{2}+10(2)+12=12+20-8=24 ft

Therefore, height after 2 s=24 ft

(b)

As it hits the ground, t is maximum and at maximum t, we differentiate the function to have

h=-4t+10  and h=0 hence

4t=10 hence t=10/4=2.5 seconds

Answer 2

Answer: A = 24ft

B = 2.5s

Step-by-step explanation:

(A)

h(t) = -2t² + 10t +12

At t = 2s

h = -2(2)² + 10(2) + 12

h = -8 + 20 + 12

h = 24ft

(B)

When will the ball hit the ground?

When t is at maximum, h = 0

0 = -2t² + 10t +12

2t² - 10t + 12 = 0

t² - 5t + 6 = 0

Solving the quadratic equation, either by factorization or formula method,

t = 2.5s