A man stands on the roof of a building of height 15.5m and throws a rock with a

Question

A man stands on the roof of a building of height 15.5m and throws a rock with a velocity of magnitude 29.6m/s at an angle of 27.9? above the horizontal. You can ignore air resistance.

Please shoe the steps to get the answers along with the answers to get the best answer.

Part A

Calculate the maximum height above the roof reached by the rock.

Part B

Calculate the magnitude of the velocity of the rock just before it strikes the ground.

Part C

Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

Explanation / Answer

Voy = Vo*sin27.9 = 29.6*sin27.9 = 13.85 m/s

Vox = Vo*cos27.9 = 26.15 m/s

A)

H = Voy^2/2g = (13.85*13.85)/(2*9.8) = 9.78 m

B) Vy ^2 - voy^2 = 2*a*H

Vy^2 - (13.85*13.85) = 2*9.8*9.78 = 19.58 m/s

V = sqrt(Vx^2+Vy^2) = sqrt(19.58^2+26.15^2) = 32.67 m/s

Vy = Voy-gT

T = 3.4

X =Vx*T = 88.91m

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