Throwing a stone weighing 1 kg, the boy applied a force of 40N on the path of 0.5m

Throwing a stone weighing 1 kg, the boy applied a force of 40N on the path of 0.5m. To what height did the stone rise after being detached from the palm?

Problem data: m (weight of the stone thrown by the boy) = 1 kg; F (force applied to the stone) = 40 N; S (the path that the stone traveled in the boy’s palm) = 0.5 m.

Constants: g (acceleration due to gravity) ≈ 10 m / s2.

We express the height to which the stone has risen after the separation from the boy’s palm from the equality: F * S = A = ΔEn = m * g * h, whence h = F * S / (m * g).

Let’s make a calculation: h = 40 * 0.5 / (1 * 10) = 2 m.

Answer: After breaking away from the boy’s palm, the stone rose 2 meters.