Posted by **Cassandra** on Tuesday, September 8, 2009 at 3:41am.

A model rocket blasts off and moves upward with an acceleration of 12 m/s^2 until it reaches a height of 28 m , at which point its engine shuts off and it continues its flight in free fall.

What is the speed of the rocket just before it hits the ground?

- Math -
**MathMate**, Tuesday, September 8, 2009 at 6:46am
So there are two steps since two accelerations are involved.

Step 1: acceleration = +12 m/s/s

initial velocity, u=0

final velocity = v

distance travelled, S=28 m

using

V²-u²=2aS

we find

v²

=2aS+u²

=2*12*28+0

=4√42 m/s (upwards)

=25.92 m/s (upwards)

Step 2: free fall, a=-9.81 m/s/s

initial velocity, u = 4√42 m/s

distance travelled, S = -28 m

final velocity = v

Again, we use the formula

V²-u²=2aS

from which

v²

=2aS+u²

=2*(-9.81)*(-28)+(4√42)²

=1221.36

v=√1221.36

=34.95 m/s (downwards)

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