Solution: A rocket is fired at an angle from the top of a tower of height exttip{h_0}{h0} = 50.1 m . Because of the design of the engines, its position coordinates are of the form x( t ) = A + Bt2 and y( t ) =

A rocket is fired at an angle from the top of a tower of height exttip{h_0}{h0} = 50.1 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 1.60 s after firing is
a = ( 4.10 i+ 3.60 j)².
Take the origin of coordinates to be at the base of the tower.

Find the constant A.

At the instant after the rocket is fired, what is its acceleration vector?

What are the x- and y-components of the rockets velocity 15.1 s after it is fired?

What is the position vector of the rocket 15.1 s after it is fired?

Find the constant B.

Find the constant C.

Find the constant D.

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

How fast is it moving 15.1 s after it is fired?