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

Question

A rocket is launched at 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 3.00s. 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 (hmax) reached by the rocket.

C) Find the flight-time when the rocket is in projectile motion.

D) Solve for the total groud distance d.

Explanation / Answer

Vf = Vi + a ? t ? y = Vi ? t + 0.5a( ? t)^2 (Vf)^2 = (Vi)^2 + 2a ? y 3. The attempt at a solution For Part A I calculated displacement with ? y = Vi ? t + 0.5a( ? t)^2. y was my unknown variable and my Vi was 0, my t 10 seconds, and my a +4. I got 200meters. For Part B I calculated the first stage's final speed by using the equation Vf = Vi + a ? t. I had a Vi of 0, an accleration of +4 and a t of 10. My velocity final was +40 m/s. As far as part c goes I plugged everything into (Vf)^2 = (Vi)^2 + 2a

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