Calculus Help!!?

Question

Please help me solve this problem, I have no idea how to start it.

A rock thrown vertically upward from the surface of the moon at a velocity of 24 lm/sec reaches a height of s= 24t - 0.8t² meters in t seconds.

a) Find the rock's veloctiy and acceleration (the acceleration in this case is the acceleration of gravity on the moon.)

b) How long did it take the rock to reach it's highest point?

c) How high did the rock go?

d) How long did it take the rock to reach half of its maximum height?

e) How long was the rock aloft?

If you could show your steps and explain what you are doing it would help alot. Thanks for your help!

Answer 1

a) s = 24t - 0.8t^2 is a form of the familiar y = (v_0)t - 0.5*gt^2, where v_0 is the initial upwards velocity and g is the acceleration due to gravity. Here, you can see that v = 24 and 0.5*g = 0.8; you can solve algebraically for g, the acceleration. For velocity, use v = v_0 - gt, where you know v_0 and g.

b) The rock reaches its highest point when its velocity is zero, right before it starts falling downward instead. That occurs at t = v_0 / g.

c) Use the equation you were given, s = 24t - 0.8t^2, to solve for s at the t you found in b).

d) Take the s_max you found in c) and solve s_max / 2 = 24t - 0.8t^2, where t is your only unknown.

e) The rock takes as much time to fall down from its highest point as it did to reach it in the first place. Just double the t you found in b). Or else solve s = 24t - 0.8t^2 for s = 0; one solution will be s = 0, when you threw it in the first place, but the other solution will be s > 0, the instant the rock hits the ground and again has a height of zero.