Physics 107 Homework help (velocity)?

Question

A model rocket fired vertically from the ground ascends with a constant vertical acceleration of 49.0 m/s2 for 1.21 s. Its fuel is then exhausted, so it continues upward as a free-fall particle and then falls back down. (a) What is the maximum altitude reached? (b) What is the total time elapsed from takeoff until the rocket strikes the ground?

I tried this question and according to WileyPlus, the website the homework is on, it's incorrect. Can someone point me in the right direction on how to solve it?

The picture I attached is the work I did trying to solve the problem.

electron1 · Answer

The first step is to use the following equation to determine the rocket’s velocity at 1.21 seconds.

vf = vi + a * t, vi = 0
vf = 49 * 1.21 = 59.29 m/s. Next use the following equation to determine its height at 1.21 seconds.

h = ½ * (vi + vf) * t
h = ½ * 59.29 * 1.21 = 35.87045 meters

As the rocket rises from this position to its maximum height, its velocity will decrease from 59.29 m/s to 0 m/s at the rate of 9.8 m/s each second. Use the following equation to determine the time for this to happen.

vf = vi – a * t, a = 9.8
0 = 59.29 – 9.8 * t 
t = 59.29 ÷ 9.8 = 6.05 seconds. Use the following equation to determine the distance it rises.

d = ½ * (vi +vf)* t, vf = 0
d = ½ * 59.29 * 6.05 = 179.35225 meters
The maximum height is the sum of these two distances.
Maximum height = 35.87045 + 179.35225 = 215.2227 meters
Time = 1.21 + 6.05 = 7.26 second

As the rocket falls 215.2227 meters, its velocity increases at the rate of 9.8 m/s each second. To determine the time, use the following equation.

d = vi * t + ½ * a * t^2, vi = 0
215.2227 = ½ * 9.8 * t^2
4.9 * t^2 = 215.2227
t^2 = 215.2227 ÷ 49 = 4.3923
t = √4.3923
This is approximately 2.096 seconds.

Total time = 7.26 + √4.3923
This is approximately 9.36 seconds.

Trending News

Physics 107 Homework help (velocity)?

1 Answer