Question

a paintball gun supplies a force of 65 N on a 0.024kg
pellet for 0.41m. ignoring the effects of air resistance, how high
would the pellet reach if it was shot straight forward?

please show work

Answer #1

**Answer:**

Given, Force F = 65 N, mass of the pellet m = 0.024 Kg and distance the pellet experienced dl = 0.41m.

When a paintball gun supplies a force F on a pellet m for a distance dl, there is an acceleration and work done takes place.

Work done W = F.dl = (65 N) (0.41m) = 26.65 J. The work will
convert into kinetic energy K = mv^{2}/2. i.e., W = K.

and then at the point of maximum height K will convert into Potential energy U, that means at maximum height, kinetic energy K = 0 and U = mgh becomes maximum.

Therfore, W = U = mgh, which implies that height h = U/mg =
(26.65 J) / (0.024 kg) (9.8 m/s^{2}) = **113.3
m.**

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