A rocket moves upward, starting from rest with an acceleration of 32.0 m/s2 for 3.27 s. It runs out of fuel at the end of the 3.27 s but does not stop. How high does it rise above the ground?
work this in two parts.
launch vf(at end of burn)=a*t
heightatendofburn;1/2 a t^2
Now the second part, start at the height above, with a vi = vf of the burn.
find the max height. at the top, vf=0
vf^2=0=vi^2+2ah where g=-9.8m/s^2, solve for h. Add that h to the heightattheendof burn
To determine the height the rocket rises above the ground, we can use the kinematic equation:
h = h0 + v0t + (1/2)at^2
Where:
h = final height above the ground
h0 = initial height above the ground (assuming it is at ground level)
v0 = initial velocity (which is 0 m/s since it starts from rest)
t = time interval (3.27 s in this case)
a = acceleration (32.0 m/s^2)
Since the rocket starts from rest, h0 = 0. Therefore, the equation simplifies to:
h = v0t + (1/2)at^2
Substituting the given values:
h = (0)(3.27) + (1/2)(32.0)(3.27)^2
Simplifying further:
h = 0 + 0.5 * 32.0 * (3.27)^2
h = 0 + 0.5 * 32.0 * 10.6929
h = 0 + 0.5 * 342.1728
h = 0 + 171.0864
Therefore, the rocket rises approximately 171.1 meters above the ground.
To find the height the rocket rises above the ground, we can use the equations of motion.
The equation we will use is:
s = ut + (1/2)at^2
Where:
s = displacement or height
u = initial velocity (which is 0 m/s since the rocket starts from rest)
t = time (3.27 s)
a = acceleration (32.0 m/s^2)
Substituting the values into the equation, we get:
s = 0(3.27) + (1/2)(32.0)(3.27)^2
Simplifying:
s = 0 + 0.5(32.0)(10.6929)
Calculating:
s = 0 + 0.5(342.85)
s = 0 + 171.43
s = 171.43 meters
Therefore, the rocket rises to a height of 171.43 meters above the ground.