A rocket blasts off and moves straight upward from the launch pad with constant acceleration. After 2.7 s the rocket is at a height of 93 m. What are the magnitude and direction of the rocket's acceleration?
What is its speed at this elevation?

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

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

the ima (higher than 1m) will be greater than the ima with the height of 1m.