Question

A rocket is launched from the top of a 40 foot cliff with an initial velocity of 150 feet per second. The height, h, of the

ocket after t seconds is given by the equation h= -16t² + 150t+ 40. How long after the rocket is launched will it be 10
feet from the ground?

Answer 1

Answer: To find out how long after the rocket is launched will it be 10 feet from the ground, we need to solve for t in the equation:

h = -16t^2 + 150t + 40

We know that when the rocket is 10 feet from the ground, h = 10:

10 = -16t^2 + 150t + 40

Subtracting 10 from both sides:

0 = -16t^2 + 150t + 30

Dividing both sides by -2:

0 = 8t^2 - 75t - 15

Using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 8, b = -75, and c = -15.

Plugging in these values:

t = (-(-75) ± sqrt((-75)^2 - 4(8)(-15))) / 2(8)

Simplifying:

t = (75 ± sqrt(5715)) / 16

Therefore, the rocket will be 10 feet from the ground approximately 0.26 seconds and 8.05 seconds after it is launched.

Your welcome (:

Explanation:

Answer 2

42

Explanation:

42-7=69z3