A stone thrown horizontally from the top of a vertical cliff with velocity 15 m s-1 is observed to strike the (horizontal) ground at a distance of 45 m from the base of the cliff. What is (a) the height of the cliff, (b) the angle the path of the stone makes with the ground at the moment of impact?
Answers: (a) 45 m, (b) 63.4°
The Attempt at a Solution
I used s = v*t to find t: 45 m = 15 m s-1 * t -> t = 3 s
Then I used s = 1/2 (u + v)*t -> s = 1/2 (15 m s-1 + 15 m s-1)*3 s = 45 m. I guess velocity when t = 0 (u) should be also 15 m s-1, equal to v = 15 m s-1. But I am not sure about the solution.
For the angle if height and horizontal distance both are 45 meters then tan angle = 45/45 -> angle = 45. But it's not the book answer.
Any help please?