Ask Question
5 April, 19:03

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?

+4
Answers (1)
  1. 5 April, 22:54
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “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 ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers