A man stands on the roof of a 19.0m -tall building and throws a rock with a velo

Question

A man stands on the roof of a 19.0m -tall building and throws a rock with a velocity of magnitude 24.0m/s at an angle of 38.0 above the horizontal. You can ignore air resistance.

a) Calculate the maximum height above the roof reached by the rock. --> already calculated it is 11

b. Calculate the magnitude of the velocity of the rock just before it strikes the ground. ( answer in m/s)

c. Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground. (answer in m)

please explain very lost on part B and C!

Thanks!!!

Explanation / Answer

Vox = 24*cos38 = 18.9 m/s

Voy = 24*sin38 = 14.78 m/s

Vy^2 - Voy^2 = 2gh

Vy^2 - 14.78^2 = 2*9.8*19

Vy = 24.3 m/s

V = sqrt(Vox^2+Vy^2) = 30.79 m/s

C) Vy = Voy - gT

-24.3 = 14.78 - 9.8*T

T = 3.98s

X = Vx*T = 75.22m

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