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Kinematics Question
By Chuck Baker on Saturday, October 27, 2001 - 07:36 pm:

During a test a rocket is traveling upward at 75 m/s, and when it is 40 m from the ground its engine fails. I want to determine the maximum height, h, reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81 m/s2 due to gravity.

We're assuming that the mass of the rocket doesn't change, in other words it's not burning any propellant. If it did, you'd have to
use the rocket equation, but since we're ignoring that, you can get away with using only ballistic equations.

During the first 40 meters of ascent, the rocket travels at a constant velocity of 75 m/s, because the engine thrust exactly balances the force of gravity, resulting in no net acceleration. But at 40 meters, the engine fails, and at this point, the rocket becomes a pure ballistic object subject only to gravitational acceleration.

In ballistics, you assume constant acceleration:

a = g

where g = -9.81 m/s2. Integrating with respect to time gives an equation
for velocity:

vf = g × t + vi

where t is time, vf is the final velocity and vi is the initial velocity.
Integrate again to get an equation for position:

yf = (1/2) × g × t2 + vo × t + yi

where yf is the final position and yi is the initial position. If we solve
the velocity equation for time:

t = (vf - vi) / g

But we know that the final velocity of the rocket is 0:

t = -vi / g

Now plug in this value for time into the equation for position:

vi2
yf = ------ + yi
-2×g

Plugging in 75 m/s for vi, 40 m for yi, and -9.81 m/s2 for g gives yf = 326.7 meters apex height.

How do you determine the impact velocity?

By George Walker on Saturday, October 27, 2001 - 09:37 pm:

There is a much simpler way of doing this problem as the acceleration is constant. We can use the kinematics equations which are used as standard results:

so for maximum height:

u=75
a=-9.81
v=0
s=?

v2 = u2 + 2as
0 = 5625 - 19.62 s

we add 40 metres to s as that is how far above the ground it started,
40 + s = 326.7m

for impact velocity, we say:

u=75
s=-40
a=-9.81
v=?

using the same equation as above:
v2 = 5625 + 2×9.81×40
v = 80.1 m/s

hope that clears it up

George