Question

A stone of mass ‘m’ is to be thrown to a height h

1. What is the acceleration of the stone?

2. With what minimum velocity should it be thrown.

3. At what height does the KE and PE become equal?

4. Find the velocity at that height

Answer 1

1. ^–g or ^–9.8m/s.

2. v = 0, a = -g, S = h

Substitute this values in

V² = u² + 2as we get

0 = u² – 2gh

u = \(\sqrt{2gh}\)

3. at \(\frac{h}{2}\)., KE and PE are equal.

4. V² = U² + 2aS

= U² – 2g \(\frac{h}{2}\) (u² = 2gh) = U² – \(\frac{U^2}{2}\),

V² = \(\frac{U^2}{2}\),

V = \(\frac{U}{√2}\).