Question

a model rocket rises with constant accerleration to a height of 3.2 m, at which point its speed is 26.0 m/s. (a) how much time does it take for the rocket to reach this height? (b) what was the magnitude of the rocket's acceration? (c) find the height and speed of the rocket 0.10 s after launch.

Answer 1

The equations that apply to rising bodies are
.....Vf = Vo - gt (the term “g” for acceleration due to gravity is assumed constant on, or near, the surface of the Earth)
.....d = Vo(t) - g(t^2)/2
.................2
.....Vf^2 = Vo^2 - 2gd

From Vf^2 = Vo^2 - 2gd
.....26^2 = 0^2 + 2(a-9.8)3.2 from which
.....a = 115.425m/s^2

From Vf = Vo - gt
.....26 = 0 + (a-9.8)t from which
t = .246sec.

At t = .10sec, from
Vf = Vo - gt
Vf = 0 + (a-9.8).10 From which
Vf = 10.56m/s

At t = .10sec, from
d = Vo(t) - g(t^2)/2 from which
d = o(.10) + (a-9.8)(.10)^2/2 from which
d = .528met.