Question

2. A stone is thrown vertically upward with a speed of 22m/s.
a. How fast is it going when it reaches a height of 25m?

Answer 1

Answer:

Explanation:

Energy E is conserved:

$E=\frac{1}{2}mv^2+mgh$

If v₀ = 22m/s, h₀=0m and h₁=25m:

$E=\frac{1}{2}mv_0^2=\frac{1}{2}mv_1^2+mgh_1$

Solving for v₁:

$v_1=\sqrt{v_0^2-2gh_1}$

There is no real solution, because the stone never reaches 25m.