A man stand on the roof of a building that is 30.0 m tall and throws a rock with

Question

A man stand on the roof of a building that is 30.0 m tall and throws a rock with a velocity of magnitude 40.0 m/s at an angle of 33.0 degrees above the horizontal.Air resistance can be ignored.Calculate:
A) The maximum height above the roof reached by the rock.
B) the magnitude of the velocity of the rock just before it strikes the ground.
C) the horizontal distance from the base of the building to the point where the rock strikes the ground.

Explanation / Answer

a) v_y = 40 * sin33 = 21.786 m/s y = (v_y)^2/(2g) = (21.786*21.786 )/(2*9.8) = 24.2 m b) v_x = 40 * cos33 = 33.547 m/s v_y = (2*9.8*(30+24.2158))^0.5 = 32.598 >>> v = (33.547*33.547+32.598*32.598)^0.5 = 46.8 m/s c) t = v_f - v_i/g = (32.598 - (-21.786))/9.8 = 5.5494 s D = v_x t = 33.547 * 5.5494 = 186 m

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