What is the position vector of the rocket

Reference no: EM13709317

Solve the all parts of given question:

Question: A rocket is fired at an angle from the top of a tower of height h0 = 55.7m. 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 1.30s after firing is a?=( 3.10 i^+ 3.00 j^)m/s². Take the origin of coordinates to be at the base of the tower.

Part A: What are the x- and y-components of the rocket's velocity 15.0s after it is fired?

Part B: How fast is it moving 15.0s after it is fired?

Part C: What is the position vector of the rocket 15.0s after it is fired?

You must show your work in an efficient manner so I can understand it in clearly.

Reference no: EM13709317

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