Thrown stone physics

Question

A stone thrown horizontally from a height of 6.12 m hits the ground at a distance of 13.30 m. Calculate the initial speed of the stone. Neglect air resistance.

Calculate the speed of the stone as it hits the ground. Neglect air resistance.

Answer 1

1. First find the time that the stone is in the air. Remember that we keep vertical and horizontal motion separate.
2. Vertical
3. The stone will hit the ground in the same time as if you had just dropped this stone from 6.12 m. Your acceleration is g and your initial vertical speed is 0 m/s
4. d = v_i * t + 1/2 * a * t²
5. Solve for t.
6. Horizontal
7. We now know the time and the distance, so we can use v = d/t to find the initial speed, since horizontal speed stays constant with projectiles without air resistance.
8. Part B:
9. We just found the horizontal component of the speed of the stone when it hits the ground. We will combine that with the vertical speed to calculate the magnitude of the velocity vector, which I guess your teacher is just calling "speed".
10. For the vertical motion, we basically know everything about it except the final speed. There are at least two 1-d motion equations that you can use to solve for v_f
11. To combine these two components into one speed, we can use the Pythagorean theorem.
12. It might help to draw out this right triangle, with the sides being the horizontal and vertical speeds you've found, and the diagonal being the final speed that you're solving for.
13. v_x² + v_y² = v_total²
14. Solve for v_total and you have your answer.