# physics

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A model rocket is launched straight upward with an initial speed of 30.0 m/s. It accelerates with a constant upward acceleration of 2.50 m/s2 until its engines stop at an altitude of 170 m.
(a) What is the maximum height reached by the rocket?
m

(b) How long after lift-off does the rocket reach its maximum height?
s

(c) How long is the rocket in the air?
s

• physics -

Maximum height occurs where the velocity is temporarily zero.

While accelerating, the height is
y(t) = 30 t + 1.25 t^2
and the speed is
v(t) = 30 + 2.5 t

First figure out when y = 170m, and what the speed is at that time.

1.25 t^2 +30t -170 = 0
t^2 + 24t -136 = 0
That does not factor easily, so use the quadratic formula.
t = (1/2)(-24 + 33.47) = 4.73 s
v(t=4.73) = 30 + 11.83 = 41.83 m/s

Zero velocity will be reached after an additional time t', such that
g*t' = 41.83 m/s
t' = 4.27 s.
The total time after liftoff is then 9.00 seconds
The maximum altitude reached is
170 m + 41.83 t' -(g/2)t'^2
= 170 + 178.6 - 89.3 = 259.3 m

(b) t + t' = 9.00 s

(c) 9.00 s + (time to fall from 259.3 m)

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