Math
posted by Cassandra .
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?

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
V²u²=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
V²u²=2aS
from which
v²
=2aS+u²
=2*(9.81)*(28)+(4√42)²
=1221.36
v=√1221.36
=34.95 m/s (downwards)