A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.30 . At 10.0 after blastoff, the engines suddenly fail, which means that the force they produce instantly stops.
How high above the launch pad will the rocket eventually go?
Find the rocket's velocity at its highest point.
Find the magnitude of the rocket's acceleration at its highest point.
Find the direction of the rocket's acceleration at its highest point. (horizontal, downward, upward)
How long after it was launched will the rocket fall back to the launch pad?
How fast will it be moving when it does so?
To find the answers to these questions, we can use the equations of motion. Let's break down each question and step-by-step explain how to find the answer:
1. How high above the launch pad will the rocket eventually go?
To find the maximum height reached by the rocket, we can use the kinematic equation:
s = ut + 0.5at^2
Where:
s = displacement (height above the launch pad)
u = initial velocity (0 m/s since it started from rest)
t = time elapsed (10.0 s)
a = acceleration (2.30 m/s^2)
Plugging in the values, we get:
s = 0(10.0) + 0.5(2.30)(10.0)^2
s = 0 + 0.5(2.30)(100.0)
s = 0 + 115.0
s = 115.0 m
Therefore, the rocket will eventually go 115.0 meters above the launch pad.
2. Find the rocket's velocity at its highest point.
To find the velocity at the highest point, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (what we want to find)
u = initial velocity (0 m/s)
a = acceleration (-2.30 m/s^2, negative because it is in the opposite direction of initial acceleration)
s = displacement (height above the launch pad, which we found as 115.0 m)
Plugging in the values, we get:
v^2 = 0^2 + 2(-2.30)(115.0)
v^2 = 0 + (-5.28)(115.0)
v^2 = -607.2
v = √(-607.2)
The square root of a negative number is not a real value, which means the rocket will not reach its highest point and will start falling before that point.
3. Find the magnitude of the rocket's acceleration at its highest point.
Since the rocket will not reach its highest point, there won't be a separate magnitude of acceleration at that point.
4. Find the direction of the rocket's acceleration at its highest point. (horizontal, downward, upward)
Since the rocket will not reach its highest point, there won't be a separate direction of acceleration at that point.
5. How long after it was launched will the rocket fall back to the launch pad?
To find the time it takes for the rocket to fall back to the launch pad, we can use the equation of motion:
s = ut + 0.5gt^2
Where:
s = displacement (height above the launch pad, which is 115.0 m)
u = initial velocity (0 m/s since it started from rest)
t = time elapsed (what we want to find)
g = acceleration due to gravity (approximately 9.8 m/s^2, assuming no significant air resistance)
Plugging in the values and rearranging the equation, we get:
115.0 = 0(𝑡) + (0.5)(9.8)(𝑡)^2
115.0 = 4.9𝑡^2
𝑡^2 = 115.0/4.9
𝑡^2 = 23.47
Taking the square root of both sides, we get:
t = √23.47
Therefore, the rocket will fall back to the launch pad approximately 4.85 seconds after it was launched.
6. How fast will it be moving when it does so?
To find the velocity at the time the rocket falls back to the launch pad, we can use the equation of motion:
v = u + gt
Where:
v = final velocity (what we want to find)
u = initial velocity (0 m/s)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time (which we found as approximately 4.85 s)
Plugging in the values, we get:
v = 0 + (9.8)(4.85)
v = 47.53 m/s
Therefore, the rocket will be moving at a speed of approximately 47.53 m/s when it falls back to the launch pad.