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 equations of motion. Since the rocket starts from rest, its initial velocity (u) is 0 m/s.
We can use the equation: s = ut + (1/2)at^2
Where:
s is the distance traveled (height above the ground),
u is the initial velocity (0 m/s),
a is the acceleration (35.0 m/s^2),
and t is the time interval (5.63 s).
Plugging in the values, we have:
s = (0)(5.63) + (1/2)(35.0)(5.63)^2
First, let's calculate (1/2)(35.0)(5.63)^2:
(1/2)(35.0)(5.63)^2 = 0.5 * 35.0 * 5.63 * 5.63 = 559.7375
Now, let's calculate s:
s = 0 + 559.7375 = 559.7375 m
Therefore, the rocket rises to a height of 559.7375 meters above the ground.