At time t=0, an astronaut stands on a platform 3 meters above the moon's surface and throws a rock directly upward with an initial velocity of 32 m/sec. if the acceleration due to gravity is 1.6 m/sec^2.

Question

a) find the equations of motion for the throwing of this rock.
-1.
a(t)=-1.6 v(t)= -1.6t + 32 s(t)=-0.8t^2 + 32t +3

b) how high above the surface of the moon will the rock travel?

the rock traveled 323 m.
c) approximately how long is the rock in the air?

t= 40.0935 s

My teacher told me that I have to get these answers for this question but I don"t know how to solve this.

Show work please! Thanks!

Answer 1

F = m a
the only force is g = -1.6 m
so
F = - 1.6 m = m a
so
a = -1.6 m/s^2

v = integral a dt = -1.6 t + constant
when t = 0, v = 32
so
v = -1.6 t + 32

h = integral v dt
h = -(1.6/2) t^2 + 32 t + constant
when t = 0 , h = 3
so
h = -0.8 t^2 + 32 t + 3

what is max h?
answer: when v = 0 No speed up, none down, at top
0 = -1.6 t + 32
t = 20 seconds
h max = -.8(400) + 32(20) + 3
h max = 323 meters

when does h = 0? (ground or moon )
-0.8 t^2 + 32 t + 3 = 0

t = {-32 +/- sqrt(1024 + 9.6) ]/-1.6

use + time
t = 40.0935313 seconds