A man stands on the roof of a 22.0-m tall building and throws a rock with a velo

Question

A man stands on the roof of a 22.0-m tall building and throws a rock with a velocity of magnitude 25.0 m/s at an angle of 35.0 degree above the horizontal. You can ignore air resistance. Calculate the maximum height above the roof reached by the rock. Calculate the magnitude of the velocity of the rock just before it strikes the ground. Calculate the horizontal range from the base of the building to the point where the rock strikes the ground. Draw x - t, y - t, v_x - t and v_y - t graphs for the motion. (Do this on paper. Your instructor may ask you to turn in these graphs.)

Explanation / Answer

Vy(0) = 25*sin35 = 12.5m/s vertically up
Vx(0) = 25*cos35 = 4.162horizontally
Using Vf² = Vi² + 2*a*d where Vf = 0 at max height, Vi = Vy(0), a = -9.8 and d = maximum height.
0 = Vy(0)² - 2*9.8*d => d = Vy(0)²/19.6 = 13.7m
1)Therefore the height above the ground at max height = 12.5+4.16 = 16.66m
The rock falls from 16.66 from a stop so we can write:
16.66 = 1/2 *g*t^2 => t = 2.483s from maximum height to impact
2)Therefore at impact Vy(t) = g*t = 9.8*2.483s = 24.3m/s down
We have to find the total time of flight. By symmetry, the time to max height from launch = the time from max height back to launch height. How long does it take to reach max height?
Vy(t) = Vy(0) - gt = 0 when t = Vy(0)/g at max height = 12.5/9.8 = 1.27 s
Therefore the TOTAL flight time = 1.27 + 2.483 = 3.753 s
Therefore the range R = Vx(0) 3.753s = 26.4*3.753 = 99.07m

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