Question

8. Given the equation h(0) = vt + ļat where h() is the height above the ground of an object

launched from the ground at time t with an initial velocity v and a constant acceleration a
(3 points total)
a. Write the function for an object with an initial velocity of 15 meters per second and a
constant acceleration of -9.8 m/s2
b. How long does it take the object to return to the ground? Show your work or explain
your response.

Answer 1

3.9 s

Step-by-step explanation:

The formula for the height H of a projectile above the ground after time t

H = (−1/2)gt² + vt + h

where g is the acceleration due to gravity, which on Earth is 32 ft/s²

v is the initial velocity = 62 ft/s

h is the object's starting height above ground = 0 ft (since our rocket is launched from the ground whose height is 0 in this question)

H = (−1/2)(32)t² + 62t + 0

H = 62t - 16t²

How long does it take until the rocket returns to the ground?

When the rocket returns back to the ground, H = 0

H = (−1/2)gt²+vt+h

0 = (-1/2)(32)t² + 62t + 0

(62t - 16t²) = 0

(16t² - 62t) = 0

t(16t - 62) = 0

t = 0 s or (16t - 62) = 0

t = 0 s or t = (62/16) = 3.875 s

t = 0 s or t = 3.9 s

Since the rocket starts from rest at t = 0, the only feasible answer has to be t = 3.9 s

Explanation: