A rocket\'s trajectory is shown in the figure attached. Details and state of mot

Question

A rocket's trajectory is shown in the figure attached. Details and state of motion after 10 seconds from its' launching point A (@ time t = 0) are given below:

r = 2200m, v = 500m/s, a = 4.66m/s^2

angular displacement = 22 degrees, angular velocity = 0.0788 rad/s, angular accel = -0.0341 rad/s^2.

For this instant, determine the angle (beta on the diagram) between the horizontal and the direction of the rocket's trajectory, and calculate the magnitude of the velocity and the acceleration of the rocket.

Also, it is known that at this same instant that the rocket detaches its' stage 1 booster. Calculate the horizontal distance from the launch pad A to the 1st rocket engine module when it lands on the earth's surface (assuming that both A and the point the engine module lands are on flat, level ground).

Explanation / Answer

x=rsin=824.13m

y=rcos=2039.80

vx=vcos+rsin

vy=vsin-rcos

ax=acos+rsin

ay=asin-rcos

t=10s

vx=vox+axt ...(1)

vy=voy+ayt ...(2)

x=rsin=voxt+0.5axt2 ...(3)

y=rcos=voyt+0.5(ay-g)t2 ...(4)

in these 4 eqns vox,voy(initial velocities)and are unknown.

solving them we get ,vox,voy.

tan=vy/vx

vtotal=(v2y+v2x)

atotal=(a2y+a2x)

for booster

x=vxt`+0.5axt`2 ...(5)

y=0=vyt`+0.5(ay-g)t`2 ...(6)

t`=time taken by booster to reach ground after it is detached.

from eqn (6) we get t`

put it in eqn (5) to get x

xtotal=x(from eqn (5)) +824.13

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