A rock is thrown off a 41m cliff at 19 m/s. It was thrown at a 63° above the ground. How long will the rock be in the air before it hits the ground?
How far from the base of the cliff with the rock land?

You can consult the previous question of yours that I answered [19947994] for the first part of this one. Just swap out the starting height, velocity, and launch angle. You would get an airtime of about t ≈ 5.097 s.

For this problem, the rock travels a horizontal distance x after time t according to

x = (19 m/s) cos(63°) t

Plug in the airtime:

x = (19 m/s) cos(63°) (5.097 s) ≈ 43.96 m ≈ 44 m

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