A rocket is launched a t an angle of 53.0 degrees to the horizontal at 360 km/h

Question

A rocket is launched a t an angle of 53.0 degrees to the horizontal at 360 km/h and moves along a straight line with a total constant acceleration of 30.0 m/s^2 for 3.00 s. At the end of the third second the rocket's engine turns off and it proceeds to move in projectile motion.
a) determine the rocket's displacement and velocity (in terms of the units vectors i and j) at 3.00 s
b) obtain the maximum height reached by the rocket.
c) find the flight time when the rocket is in projectile motion
d) solve for the total ground distance d.

Explanation / Answer

The distance covered by the rocket in 10 sec d1 = Vi*t + 1/2 at^2 = 0 + 0.5*49*10^2 = 2450 m The speed after 10 sec V = Vi + a*t = 0 + 49*10 = 490 m/s At the highest point Vf = 0. Therefore, distance covered under free fall can be determined by Vf^2 = V^2 + 2a*d2 Or 0 = 490^2 - 2*9.8*d2 Or d2 = 12250 m Thus maximum height = d1 + d2 = 2450 + 12250 = 14700 m

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