Find its height and velocity 10 seconds later

Question

On the surface of the moon, the acceleration of gravity is -5.28 feet per second per second. If an object is thrown upward from an initial height of 1000 feet with a velocity of 20 feet per second, find its height and velocity 10 seconds later.

Answer 1

Acceleration a = -5.28

Integrate the acceleration to obtain the velocity:

a = dv/dt = -5.28

∫ dv = ∫ -5.28 dt

v = -5.28 t + C

Initial condition t = 0, v = 20, so C = 20

Velocity v = -5.28 t + 20, evaluate that at t = 10 seconds.

Integrate the velocity to obtain the position (height):

v = dx/dt = -5.28 t + 20

∫ dx = ∫ (-5.28 t + 20) dt

x = - 5.28t²/2 + 20t + C

Initial condition t = 0, x = 1000, so C = 1000

Height x = - 5.28t²/2 + 20t + 1000, evaluate that at t = 10 seconds.