A toy rocket, launched from the ground, rises vertically with an acceleration of 32 m/s^2 for 4.1 s until its motor stops. Disregarding any air resistance, what maximum height above the ground will the rocket achieve? The acceleration of gravity is 9.8 m/s^2. Answer in units of k
Vi = a t = 32 * 4.1 = 131 m/s at burn out
h = (1/2)32 (4.1)^2 = 269 meters at burn out
then
v = Vi + a t
at top v = 0
0 = 131 - 9.8 t
t = 13.4 seconds coasting up
h = 269 + 131 (13.4) -4.9 (13.4)^2
as usual,
s = 1/2 at^2
Since g is acting downward, the net acceleration is 32-9.8 = 22.2 m/s^2 upward
s = 1/2 * 22.2 * 4.1^2 = 186.6m
Actually, the question is poorly worded, since it implies that the net acceleration is 32 m/s^2, even after including the effects of gravity. Instead, it should have been stated that the rocket motor provided the 32 m/s^2 acceleration, from which gravity's effect must be subtracted.
I think they mean total acceleration up
As usual, go with Damon. He tends to think things through more carefully.
To find the maximum height achieved by the rocket, we need to use the equations of motion.
Step 1: Determine the initial velocity (u):
The rocket is initially at rest, so the initial velocity (u) is 0 m/s.
Step 2: Determine the time taken to reach maximum height (t):
The motor stops after 4.1 seconds, so this is the time taken to reach maximum height.
Step 3: Determine the final velocity (v):
We can use the equation v = u + at, where:
v = final velocity (which is 0 m/s since the rocket stops)
u = initial velocity (0 m/s)
a = acceleration (32 m/s^2)
t = time (4.1 s)
0 = 0 + (32 * 4.1)
0 = 131.2
Step 4: Determine the displacement (s):
We can use the equation s = ut + (1/2)at^2, where:
s = displacement (maximum height)
u = initial velocity (0 m/s)
t = time (4.1 s)
a = acceleration (32 m/s^2)
s = 0(4.1) + (1/2)(32)(4.1)^2
s = 65.92
Therefore, the maximum height achieved by the rocket is 65.92 meters above the ground.