a model rocket is launched vertically at v-initial = 55m/s. It accelerates at 2.

Question

a model rocket is launched vertically at v-initial = 55m/s. It accelerates at 2.2 m/s until the engine stops at 150m above the ground. a) what is the max height? b) how long is it in the air?

then:

a a model rocket is launched vertically from the ground with an acceleration of 4m/s^2 for 6 seconds. its fuel is then exhausted. What is the max altitude? How long is it in the air?

can you please walk me through the process of solving this problem so i can understand it? I think the purpose is to show different variables being given.

Explanation / Answer

1)

vi = 55 m/s
let vf is the velocity of the rocket when engine is off.

a = 2.2 m/s

h = 150 m

a) Apply, vf^2 - vi^2 = 2*a*h

==> vf = sqrt(vi^2 + 2*a*h)

= sqrt(55^2 + 2*2.2*150)

= 60.7 m/s

Hmax = h + vf^2/(2*g)

= 150 + 60.7^2/(2*9.8)

= 338 m

b) total time of flight = (vf -vi)/a + vf/g + sqrt(2*Hmax/g)

= (60.7 - 55)/2.2 + 60.7/9.8 + sqrt(2*338/9.8)

= 17.1 s

2) velocity of rocket when fuel exhausted, v = u + a*t

= 0 + 4*6

= 24 m/s

distance travelled, h = 0.5*a*t^2

= 0.5*4*6^2

= 72 m

so, Hmax = h + v^2/(2*g)

= 72 + 24^2/(2*9.8)

= 101.4 m

total time of flight, T = 6 + v/g + sqrt(2*Hmax/g)

= 6 + 24/9.8 + sqrt(2*101.4/9.8)

= 13 s

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