Question

A rocket is fired at an angle from the top of a tower of height h0 = 68.6 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 1.10 s after firing is
a⃗ =( 2.30 i^+ 2.20 j^)m/s2.
Take the origin of coordinates to be at the base of the tower.

At the instant after the rocket is fired, what is its velocity?

Answer 1

The art of physics man....*mind blown*

Answer 2

In x v(t) = 2Bt and a(t) = 2B so B = 1.15
In y v(t) = 3Dt^2 and a(t) = 6Dt making D = .33
Then things go a bit off because the velocity at t=0 is zero in x and y and the question becomes pointless.