A rocket moves upward, starting from rest with an acceleration of 28.3 m/s2 for 5.50 s. It runs out of fuel at the end of the 5.50 s, but does not stop.

How high does it rise above the ground? Answer in units of m

h1=a*t^2 = 28.3*5.5^2=856.1 m Above gnd.

V1 = a*t = 28.3 * 5.5 = 155.7 m/s.

V = V1 + g*t = 0
155.7 - 9.8t = 0
-9.8t = -155.7
t = 15.9 s. To reach max ht.

h2 = V1*t + 0.5g*t^2
h2 = 155.7*15.9 - 4.9*15.9^2 = 1236 m
Above h1.

h = h1 + h2 = 856 + 1236 = 2092 m. Above gnd.

To find the height the rocket rises above the ground, we can use the equations of motion.

First, we need to find the final velocity of the rocket. We can use the equation:
v = u + at

Where:
v = final velocity
u = initial velocity (0 m/s as the rocket starts from rest)
a = acceleration (28.3 m/s^2)
t = time (5.50 s)

Plugging in the values, we get:
v = 0 + (28.3 m/s^2)(5.50 s)
v = 155.65 m/s

Next, we can use the equation to find the distance traveled by the rocket:
s = ut + (1/2)at^2

Where:
s = distance traveled
u = initial velocity (0 m/s)
t = time (5.50 s)
a = acceleration (28.3 m/s^2)

Plugging in the values, we get:
s = (0 m/s)(5.50 s) + (1/2)(28.3 m/s^2)(5.50 s)^2
s = 383.625 m

Therefore, the rocket rises approximately 383.625 meters above the ground.