A rocket is fired at an angle from the top of a tower of height h0 = 55.7m . Bec

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 + Bt^2 and y(t)= C + Dt^3, 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.

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

How fast is it moving 15.0s after it is fired?

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

Explanation / Answer

a)v(x)=dx/dt=2Bt

a(x)=2B

v(y)=dy/dt=3Dt^2

ay=dv(y)/dt=6Dt

ax(1.3)=2B=3.1

B=1.55

ay(1.3)=6*D*1.3=3

D=0.3846

vx(15)=2*1.55*15=46.5m/s

vy(15)=3*0.3846*15^2=259.6m/s

b)speed=sqrt(46.5^2+259.6^2)=263.7m/s

c)x=Bt^2(since it is at the origin at t=0)

=1.55*15^2=348.75m

y=55.7+Dt^3=55.7+0.3846*15^3=1353.7m

Position is (348.75,1353.7)

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