A rocket moves upward, starting from rest with an acceleration of 35.0 m/s2 for 5.63 s. It runs out of fuel at the end of the 5.63 s but does not stop. How high does it rise above the ground?
To find the height the rocket rises above the ground, we can use the kinematic equation:
s = ut + (1/2)at^2
where:
s is the displacement (height)
u is the initial velocity (starting from rest, so u = 0 m/s)
t is the time interval (t = 5.63 s)
a is the acceleration (35.0 m/s^2)
Plugging in these values into the equation:
s = 0 + (1/2)(35.0 m/s^2)(5.63 s)^2
Let's calculate this:
s = (1/2)(35.0 m/s^2)(31.6969 s^2)
Simplifying the expression:
s = 0.5 x 35.0 x 31.6969 m
Calculating the result:
s ≈ 559.705 m
Therefore, the rocket rises approximately 559.705 meters above the ground before running out of fuel.