Answer:
The magnitude and direction of the rocket acceleration is 68.89 m/s² upward.
The speed of the rocket at the given elevation is 186 m/s.
Explanation:
Given;
time to reach the given height, t = 2.7 s
height reached, h = 93 m
initial velocity of the rocket, u = 0
The magnitude and direction of the rocket acceleration is calculated as;
h = ut + ¹/₂at²
h = 0 + ¹/₂at²
h = ¹/₂at²
a = 2h / t²
a = (2 x 93) / 2.7
a = 68.89 m/s²
the direction of the acceleration is upward.
The speed at this elevation, V = u + at
V = at
V = 68.89 x 2.7
V = 186 m/s
