A rocket moves upward, starting from rest with an acceleration of 28.8 m/s2 for 4.13 s. It runs out of fuel at the end of the 4.13 s but does not stop. How high does it rise above the ground?
Acceleration?
Thrust=mass*a
But thrust*distancefiring=energy put into the rocket. How high?
energy put into rocket=mgh
solve for h.
now distance firing: d=1/2 a t^2
Compute the distance that it rises while thrusting for time t. Call it X1
X1 = (1/2) a t^2
Also compute the velocity when acceleration stops. Call it Vmax.
Vmax = a t
Compute how much farther the rocket travels up until the maximum height (and zero velocity) is reached. Call that X2.
Maximum height is reached after additional time interval
t' = Vmax.g
X2 = Vmax*t' - (g/2)t'^2
The answer is X1 + X2.
Liftoff to burnout in 4.13 sec.
Burnout velocity Vbo = 28.8(4.13) = 123.84m/s.
Burnout altitude h1 = at^2/2 - gt^2/2 = 19(4.13)^2/2 = 162m.
From Vbo to V = 0, Vf = Vbo - gt = 0 = 123.84 - 9.8t or t = 12.636 sec.
From Vbo to V = 0, h2 = 123.84(12.636) - 9.8(12.636)^2/2 = 782m.
Total height H = h1 + h2.
To find the height the rocket rises above the ground, we can use the equations of motion. We'll start by finding the velocity of the rocket at the end of the 4.13 seconds.
Using the equation:
v = u + at
where:
v = final velocity
u = initial velocity (0 m/s, since the rocket starts from rest)
a = acceleration (28.8 m/s²)
t = time (4.13 s)
Substituting the values into the equation:
v = 0 + (28.8 m/s²)(4.13 s)
v = 119.13 m/s
Now, we can find the displacement or the distance the rocket travels during the time it is accelerating.
Using the equation:
s = ut + (1/2)at²
where:
s = displacement
u = initial velocity (0 m/s)
t = time (4.13 s)
a = acceleration (28.8 m/s²)
Substituting the values into the equation:
s = (0 m/s)(4.13 s) + (1/2)(28.8 m/s²)(4.13 s)²
s = (0 m) + (1/2)(28.8 m/s²)(17.0369 s²)
s ≈ 240.97 m
Therefore, the rocket rises approximately 240.97 meters above the ground.