A uniformly accelerating rocket is found to have a velocity of 11.0 m/s when its height is 4.00 m above the ground, and 1.90 s later the rocket is at a height of 56.0 m. What is the magnitude of its acceleration?

17.23 m/s²

Explanation:

Applying the equation of motion,

Δs = ut + 1/2at² ... Equation 1

Where Δs = change in height of the rocket, u = initial velocity of the rocket, a = acceleration of the rocket, t = time

making a the subject of the equation,

a = 2 (Δs-ut) / t² ... Equation 2

Given: Δs = (56-4) m = 52 m, u = 11.0 m/s, t = 1.90 s.

Substitute into equation 2

a = 2[52 - (11*1.9) ]/1.9²

a = 2 (52-20.9) / 1.9²

a = 2 (31.1) / 3.61

a = 62.2/3.61

a = 17.23 m/s².

Thus the acceleration = 17.23 m/s²