: 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!!!

i really dont know

step-by-step explanation:

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!