A test rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 49 m and acquired a velocity of 30m/s What is the maximum height that the rocket will reach?
Aracquetball strikes aWall with aSpeed of 30 m/s and rebounds with a speed of 26m/s The collision takes20milisec What is theAverage cceleration of the ball during the collision?
5. So the only thing we really need is the second part of the question because we know the rocket stops accelerating at 49m, so now take the kinematic equation d = -1/2at^2 + vt + di, which would be -4.9t^2 + 30t + 49. Now just solve for time by dividing 30 by 9.8 to get t = 3.06. Plug that into the other equation to get -4.9(3.06)^2 + 30(3.06) + 49 = about 94.92m
6. So we need to find the difference in velocities of the two trajectories. Since the trajectories go in opposite directions we add the two speeds to get 56m/s. So the ball changes velocity by 56m/s in 20 milliseconds. So convert 20 milliseconds to seconds by dividing by 1000 to get .02 seconds. divide 56m/s by 0.02 s to get 2800m/s^2 as the average acceleration
Hope this helps
May 17th, 2015
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