putongren
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- Homework Statement
- At the base of a vertical cliff, a model rocket, starting from rest, is launched upwards at t = 0 with a time-varying acceleration given by
a(t) = A - Bt
where A and B are positive constants. Also at t = 0, a small stone is released form rest from the top of the cliff at a height h directly above the rocket. (This height h is higher than the maximum height reached by the rocket.). The stone hits the rocket at the instant when the rocket reaches its maximum height. The gravitational acceleration of magnitude g, is down dropped in terms of constants A, B and g.
- Relevant Equations
- Total Distance = Distance traveled by stone + Distance traveled by rocket
Total Distance = distance traveled by stone + distance traveled by rocket.
Total Distance = h
Distance traveled by stone = xstone = (1/2)gt2
Distance traveled by rocket:
arocket(t) = A - Bt
vrocket(t) = At - (1/2)Bt2
xrocket(t) = (1/2)At2 - (1/6)Bt3
In order to find the maximum height of the rocket, let v(t) = 0
So, 0 = At - (1/2)Bt2
0 = t(A - (1/2)Bt)
Therefore, t = (0, 2A/B)
The maximum of the rocket occurs when t = 2A/B.
Plug t = 2A / B into the Total Distance eqn:
h = (1/2)g(2A/B)2 + (1/2)A(2A/B)2 - (1/6)B(2A/B)3