A rocket is fired at an angle from the top of a tower of height h0 = 31.5 {\ m m

Question

A rocket is fired at an angle from the top of a tower of height h0 = 31.5 { m m}. Because of the design of the engines, its position coordinates are of the form x(t) = A + Bt^2 and y(t) = C + Dt^3, where A, B, C , and D are constants. Furthermore, the acceleration of the rocket 1.05 { m s} after firing is
ec a = ( 2.60 hat i+ 2.15 hat j);{ m m/s}^2.
Take the origin of coordinates to be at the base of the tower. Now we can use calculus to determine expressions for the velocity and acceleration of the rocket. v(t) = 2Bt{hat i} + 3Dt^2 {hat j} and a(t) = 2B {hat i} + 6Dt {hat j}
What is the constant A?

Explanation / Answer

0 because t=0

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