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!
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)
a) To find the equations of motion for the throwing of the rock, we need to use the kinematic equations. The kinematic equations describe the motion of an object under constant acceleration.
The key equations we need are:
1) v = u + at
2) s = ut + (1/2)at^2
3) v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
s = displacement
In this case, the initial velocity of the rock is 32 m/sec, the acceleration due to gravity is -1.6 m/sec^2 (negative because it is acting in the opposite direction of the initial velocity), and the displacement is unknown.
b) To find how high above the surface of the moon the rock will travel, we can use the kinematic equation for displacement (s). We can assume the final velocity when the rock reaches its highest point is 0.
Using equation (2), we have:
0 = 32 - 1.6t (since v = u + at)
1.6t = 32
t = 32 / 1.6
t = 20 seconds
Now, we can use equation (2) again to find the displacement:
s = (32 * 20) + (0.5 * -1.6 * (20^2))
s = 640 - 320
s = 320 meters
Therefore, the rock will travel 320 meters above the surface of the moon.
c) To find approximately how long the rock is in the air, we need to determine the total time it takes for the rock to reach its highest point and then fall back down to the moon's surface.
Using the equation v = u + at, we know that the final velocity when the rock reaches its highest point is 0 m/sec. We can set this equation up as:
0 = 32 - 1.6t
1.6t = 32
t = 32 / 1.6
t = 20 seconds
Since it takes 20 seconds for the rock to reach its highest point, it will also take 20 seconds for it to fall back down to the moon's surface. Therefore, the rock is in the air for approximately 40 seconds.