Question

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?

Answer 1

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)