: A model rocket is launched from the ground with an initial velocity of 62 feet per second. The formula for the height H of a projectile after time t is given by H=−1/2gt^2+vt+h, where g is the acceleration due to gravity, which on Earth is 32 feet per second squared in U. S. Customary units, v is the initial velocity, and h is the object's starting height above ground. How long does it take until the rocket returns to the ground? Enter the answer rounded to the nearest tenth of a second.

3.9 s

Step-by-step explanation:

The formula for the height H of a projectileabove 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

Hope this Helps!!!

step-by-step explanation:

we are given

since, we have to isolate x on anyone side

so, we can move all x terms on one side

subtract both sides by 2x

add both sides by 9

divide both sides by 4

we get

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step-by-step explanation: