Homework Help: physics

Posted by ben on Sunday, October 2, 2011 at 11:32am.

A model rocket is launched straight upward
with an initial speed of 45.4 m/s. It accelerates with a constant upward acceleration of
2.98 m/s
2
until its engines stop at an altitude
of 150 m.
What is the maximum height reached by
the rocket? The acceleration of gravity is
9.81 m/s
2
.
Answer in units of m

physics - Henry, Monday, October 3, 2011 at 8:17pm
Vf^2 = Vo^2 + 2g*d,
Vf^2 = (45.4)^2 + 19.6*150 = 5001.16,
Vf = 70.7m/s = Vo for free fall phase.

h = ho + (Vf^2 - Vo^2) / 2g,
h = 150 + (0 - (70.7)^2) / -19.6=405m.
physics - ben, Monday, October 3, 2011 at 9:00pm
When does the rocket reach maximum height?
Answer in units of s
physics - ben, Monday, October 3, 2011 at 11:13pm
How long is the rocket in the air?
Answer in units of s
physics - Henry, Wednesday, October 5, 2011 at 2:19pm
Correction to your 10-2-11,11:32am post:

The acceleration for the 1st phase should be 2.98m/s :

Vf^2 = Vo^2 + 2a*d,
Vf^2 = (45.4)^2 + 5.96*150 = 2955.16,
Vf = 54.36m/s. = Vo for free-fall phase

h = ho + (Vf^2 - Vo^2) / 2g,
h=150 + (0 - (54.36)^2) / -19.6 = 301m.

t1 = (Vf - Vo) / a,
t1 = (54.36 - 45.4) / 2.98 = 3.0s. to
reach 150m.

t2 = (0 - 54.36) / -9.8 = 5.55s To go
from 150m to max. height(301m).

t(up) = t1 + t2 = 3 + 5.55 = 8.55s

d = V0*t + 0.5g*t^2 = 301m,
0 + 4.9t^2 = 301,
t^2 = 61.4,
t(dn) = 7.84s.

T = t(up) + t(dn) = 8.55 + 7.84 = 16.4s = Time in flight.

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