Answer to Question #163673 in Other for fin

Question #163673

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

ay(t) = A − Bt

where A and B are positive constants. Also at t = 0, a small stone is released from 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 downward. You may neglect air resistance. Determine an expression for the initial height h from which the stone was dropped in terms of the constants A, B, and g. 


1
Expert's answer
2021-02-16T14:19:44-0500

"a(t)=A-Bt"

"V=\\int adt=At- \\frac{Bt^2}{2} \\implies v=0 \\space at \\space t=\\frac{2A}{B}"

"x-\\int Vdt=\\frac{At^2}{2}- \\frac{Bt^3}{6} =t^2(\\frac{A}{2}- \\frac{Bt}{6})"

Distance travelled by stone in time t, "D=0.5gt^2"

"D=0.5\\times g \\times \\frac{4A^2}{B^2}=\\frac{2A^2}{B^2}g"

But we know x+D=h

"h=\\frac{A^2}{B}(2g+\\frac{4A}{6})"

"h=\\frac{A^2}{B^2}(2g+\\frac{2A}{3})"

"h=\\frac{2A^2(3g+A)}{3B^2}"


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