A rock is thrown with a height equation of h = -16t2 + 20t + 5 (where h is the height of the rock in feet at any given time of t in seconds). Will it reach a height of 30 feet? Explain your answer.

Answer 1

h = -16t² + 20t + 5

Substitute 30 feet for h.

-16t² + 20t + 5 = 30

-16t² + 20t - 25 = 0

Use the quadratic formula to solve for t.

t = -20 ± [√20² - 4(-16)(-25)] / 2(-16) = -20 ± [√400 - 1600] / (-32) = -20 ± √-1200 / (-32)

The discriminant has a negative number inside the square root. This tells you there are no real solutions. Therefore, the rock can't reach 30 feet in the air after being thrown.

Another way to look at is by its vertex. Find t by using -b/2a. t = -20/2(-16) ≈ 0.63. Plug in t to find h.

h = -16(0.63)² + 20(0.63) + 5 ≈ 11.25.

The vertex is (0.63,11.25), which means at 0.63 seconds the rock is thrown as high as 11.25 feet.

Answer 2

The maximum height occurs when x = -20/(2(-16)) = 20/32 = 5/8

Maximum height = h(5/8) = 11.25 ft. So, the rock cannot attain a height of 30 feet.

Answer 3

h = -16t^2 + 20t + 5

substitute 30 for h and solve for t

30 = -16t^2 + 20t + 5

16t^2 - 20t + 25 = 0

(4t-5)(4t-5) = 0

t = 5/4 = 1.25 seconds.

It will reach a height of 30 feet in 1.25 seconds