A projectile of mass 0.750 kg is shot straight up with an initial speed of 18.0 m/s. How high would it reach with no air resistance? What is the average force of the air exerted on it if the projectile reaches a max height of 11.8 with air resistance?

Jun 3, 2017

The force of the air resistance is $= 2.95 N$

Explanation:

The loss in kinetic energy is $=$ the gain in potential energy

$\Delta K E = P E$

$\frac{1}{2} m {u}^{2} - \frac{1}{2} m {v}^{2} = m g h$

The initial velocity is $u = 18 m {s}^{-} 1$

The final velocity is $v = 0 m {s}^{-} 1$

The height is $= h m$

$\frac{1}{2} m {u}^{2} - 0 = m g h$

$h = {u}^{2} / \left(2 g\right) = {18}^{2} / \left(2 g\right) = 16.53 m$

There is a loss of potential energy due to air resistance

$\Delta P E = m g \left(h - 11.8\right)$

This is equal to the work done by the force of air resistance

$0.750 \cdot 9.8 \left(16.53 - 11.8\right) = R \cdot 11.8$

The force of the air resistance is $R = \frac{0.750 \cdot 9.8 \cdot 4.73}{11.8} = 2.95 N$