height = -1.6t^2 + 32t + 3
b) need the vertex of that parabola
t of vertex = -b/(2a) = -32/-32 = 1 second
height = -16(1) + 32(1) + 3 = 19 m
c) hits the moon ground when h = 0
-16t^2 + 32t + 3 = 0
t = (-32 ± √1216)/-32
= -.089 or 2.09 seconds
so , "in the air" of the moon from t = 0 to t = 2.09
or 2.09 seconds
PS. there is no air on the moon, poor wording in c)
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.
a) find the equations of motion for the throwing of this rock.
b) how high above the surface of the moon will the rock travel?
c) approximately how long is the rock in the air?
Show work please! Thanks!
1 answer
1 answer