Answer:
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
