Question

A rocket is fired at an angle from the top of a tower of height h0 = 47.1 m . Because of the design of the engines, its position coordinates are of the formx(t)=A+Bt2 and y(t)=C+Dt3, where A,B, C , and D are constants. Furthermore, the acceleration of the rocket 1.00 s after firing is
a⃗ =( 3.70 i^+ 1.10 j^)m/s2.
Take the origin of coordinates to be at the base of the tower.

1)Find the constant A.

2)Find the constant B.

3)Find the constant C.

4)Find the constant D.

5)At the instant after the rocket is fired, what is its acceleration vector? ax & ay

6)At the instant after the rocket is fired, what is its velocity? Vx &Vy

Answer 1

The origin of coordinate system is not given. I am assuming that it is at the base of the tower
1.

a)x(t) = A + Bt^2
The rocket is fired from vertically above the origin. Therefore, x = 0 at t= 0
Therefore, 0 = A + B*0^2
Or A = 0

2.

Differentiating (1) with respect to t,
dx/dt = 0 + 2Bt = 2Bt
Differentiating again with respect to t,
d2x/dt^2 = 2B
d2x/dt^2 is ax
Therefore, ax = 2B
Given ax = 3.70 m/s^2
Therefore 2B = 3.70 m/s^2
Or B = 3.70/2 m/s^2
Or B = 1.85 m/s^2

3.
y(t) = C+D(t^2)
y = 47.1m at t = 0
Therefore, 47.1m = C + D*0^2
Or C = 47.1 m

4.

Differentiating (5) with respect to t,
dy/dt = 0 + 2Dt
Differentiating again w.r.t. t,
d2y/dt^2 = 2D
But d2y/dt^2 = ay
Therefore, ay = 2D
Given ay = 1.10 m/s^2
Therefore, 2D = 1.10 m/s^2
Therefore, D = 0.55 m/s^2---------------------(8)

Ans: A = 0, B = 1.85 m/s^2, C = 47.1 m, D = 0.55 m/s^2