Arocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical acceleration of the rocket is given by ay=(2.70m/s3)t, where the +y-direction is upward. what is the speed of the rocket when it is 325 m above the surface of the earth?

Explanation:

The vertical acceleration of the rocket is given by :

Acceleration is given by,

..............(1)

Velocity is given by,

From above equation, we can find the value of t at x = 325 m

t = 8.97 s

Now put t = 8.97 s in equation (1) as :

v = 108.62 m/s

So, the speed of the rocket when it is 325 meters above the surface of earth is 108.62 m/s. Hence, this is the required solution.

double the distance = divide strength by 4

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.