Question

A model rocket is launched straight upward with an initial speed of 50.2 m/s. It accelerates with a constant upward acceleration of 2.12 m/s2 until its engines stop at an altitude of 167 m. What is the maximum height reached by the rocket?

Answer 1

h = 50.2t + (2.12-9.8)/2 t^2
= 50.2t - 3.84t^2
engine kicks out when h=157
157 = 50.2t - 3.84t^2
t = 5.18

now the velocity is 50.2 + 5.18*2.12 = 61.18

now, from that point, reset the clock to 0, and its height is ballistic:

h = 157 + 61.18t - 4.9t^2
h = 157 + t(61.18 - 4.9t)
max height reached at t = 6.24
h(6.24) =~ 348m

time from blast-off is 5.18+6.24 = 11.82 seconds

Answer 2

You didn't do your work you put in a time how am I suppose to see it is correct without work to prove it. You go to math to learn proofs so use the skill if you are going to publish things. Don't be lazy

Answer 3

The engine doesn't kick out at 157, it kicks out at 167.