A rocket is fired at an angle from the top of a tower of height h0 = 65.5 m . Because of the design of the engines, its position coordinates are of the form x(t)=A+Bt2 and y(t)=C+Dt3, where A, B, C , and D are constants. Furthermore, the acceleration of the rocket 1.50 s after firing is

a⃗ =( 3.70 i^+ 3.10 j^)m/s2.
Take the origin of coordinates to be at the base of the tower.

what are constants A, B, and C

To determine the constants A, B, and C in the position coordinates of the rocket (x(t) and y(t)), we need to analyze the information given in the problem.

Given:
- The rocket is fired from the top of a tower of height h0 = 65.5 m. This means that the initial vertical position of the rocket is h0, which corresponds to the constant C in the y(t) equation.
- The acceleration of the rocket 1.50 s after firing is given as a⃗ =( 3.70 i^+ 3.10 j^)m/s2. The acceleration of an object is the second derivative of its displacement with respect to time. Since the position coordinates are given as x(t) = A + Bt^2 and y(t) = C + Dt^3, we can differentiate twice to obtain the acceleration expressions.

Differentiating x(t) twice:
x(t) = A + Bt^2
First derivative: v(t) = dx(t)/dt = 2Bt
Second derivative: a(t) = dv(t)/dt = d^2x(t)/dt^2 = 2B

Differentiating y(t) twice:
y(t) = C + Dt^3
First derivative: v(t) = dy(t)/dt = 3Dt^2
Second derivative: a(t) = dv(t)/dt = d^2y(t)/dt^2 = 6Dt

Comparing the obtained accelerations with the given acceleration vector:
2B = 3.70 (from the x-component of the acceleration)
6D = 3.10 (from the y-component of the acceleration)

Solving these two equations for B and D:
B = 3.70/2 = 1.85
D = 3.10/6 = 0.5167

Now, we have B and D. To find the remaining constant A, we can use the initial condition. The rocket is fired from the top of the tower, so its initial position is (0, h0).
Substituting t = 0 into the x(t) equation:
x(0) = A + B(0)^2
0 = A

So, A = 0.

Therefore, the constants A, B, and C in the position coordinates are:
A = 0
B = 1.85
C = h0 = 65.5 m.