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.
-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!
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
To solve this problem, we will use the equations of motion for projectile motion.
a) The equations of motion for this scenario are as follows:
- Equation 1: a(t) = -1.6 m/s^2
- Equation 2: v(t) = -1.6t + 32 m/s
- Equation 3: s(t) = -0.8t^2 + 32t + 3 m
Equation 1 describes the acceleration of the rock, which is due to gravity and is constant at -1.6 m/s^2 (negative because it acts downward).
Equation 2 gives the velocity of the rock as a function of time. It shows that the initial upward velocity of 32 m/s decreases by 1.6 m/s each second due to the acceleration caused by gravity.
Equation 3 represents the displacement (or height) of the rock above the moon's surface as a function of time. It includes the initial height of 3 meters, the additional height due to the upward velocity, and the height due to the acceleration of gravity.
b) To find how high above the surface of the moon the rock will travel, we need to determine the maximum height reached by the rock. The maximum height is achieved when the velocity of the rock becomes zero.
Setting Equation 2 to zero:
- -1.6t + 32 = 0
Solving for t:
- -1.6t = -32
- t = -32 / -1.6
- t ≈ 20 seconds
Now, substitute this value of t into Equation 3 to find the maximum height:
- s(t) = -0.8(20)^2 + 32(20) + 3
- s(t) = -0.8(400) + 640 + 3
- s(t) = -320 + 640 + 3
- s(t) ≈ 323 meters
Therefore, the rock will travel approximately 323 meters above the surface of the moon.
c) To find the time the rock is in the air, we need to find how long it takes for the rock to reach the surface of the moon again. This occurs when the displacement (height) becomes zero.
Setting Equation 3 to zero:
- -0.8t^2 + 32t + 3 = 0
This is a quadratic equation that can be solved using various methods, such as factoring or the quadratic formula. Applying the quadratic formula:
- t = (-32 ± √(32^2 - 4(-0.8)(3))) / (2(-0.8))
Calculating this, we get:
- t = (-32 ± √(1024 + 9.6)) / (-1.6)
- t = (-32 ± √(1033.6)) / (-1.6)
Taking the positive square root and solving:
- t = (-32 + √(1033.6)) / (-1.6)
- t ≈ 40.0935 seconds
Therefore, the rock is in the air for approximately 40.0935 seconds.