A rock is projected from the edge of the top of a building with an initial velocity of 12. 2 m/s at an angle of 63° above the horizontal. The rock strikes the ground a horizontal distance of 25 m from the base of the building. Assume that the ground is level and that the side of the building is vertical. What is the time?.

The rock travels a horizontal distance x after time t of

x = (12.2 m/s) cos(63°) t

If it hits the ground 25 m away from where it was thrown, then for some time t we have

25 m = (12.2 m/s) cos(63°) t

Solve for t :

t = (25 m) / ((12.2 m/s) cos(63°))

t ≈ 4.5 s

answer; mixture;

this isn't a question, nor is it appropriate for this web site

explanation:

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