A test rocket is launched vertically from ground level (y=0m), at time t=0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 97 m and acquired a velocity of 40 m/s. The rocket continues to rise in upward flight, reaches maximum height, and falls back to the ground. What is the upward acceleration of the rocket during the burn phase?
To determine the upward acceleration of the rocket during the burn phase, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity = 40 m/s
u = initial velocity = 0 m/s (since the rocket starts from rest)
a = acceleration
s = displacement = 97 m
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Substituting the given values, we get:
a = (40^2 - 0^2) / (2 * 97)
a = 1600 / 194
a ≈ 8.25 m/s^2
Therefore, the upward acceleration of the rocket during the burn phase is approximately 8.25 m/s^2.
To find the upward acceleration of the rocket during the burn phase, we can use the equations of motion.
First, let's consider the burn phase until engine burnout. We know the initial velocity (v0) is 0 m/s (because it is launched from rest), the final velocity (v) is 40 m/s, and the displacement (y) is 97 m.
The equation we can use is:
v^2 = v0^2 + 2aΔy
where v is the final velocity, v0 is the initial velocity, a is the acceleration, and Δy is the displacement.
Plugging in the given values, we get:
40^2 = 0^2 + 2a * 97
Simplifying the equation, we have:
1600 = 194a
Now, we can solve for the acceleration (a):
a = 1600 / 194
Using a calculator, we find that the upward acceleration of the rocket during the burn phase is approximately 8.247 m/s^2.
Therefore, the upward acceleration of the rocket during the burn phase is approximately 8.247 m/s^2.
Assuming constant acceleration:
use the kinematic equation Vf^2=Vo^2+2ad
there is no Vo, so the equation becomes: Vf^2=2ad
substitute and you get: 40^2=2a(97)
solve for a: a= 8.25 m/s or 8.3 m/s if you use sig. figs.
Hope this helps!