Question

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:(a) the work done by the force of gravity on the projectile,
(b) the change in kinetic energy of the projectile since it was fired, and
(c) the final kinetic energy of the projectile. (d) Are any of the answers changed if the initial angle is changed?

Answer 1

Answers:

a) $W_{g}=mgh$

b) $\Delta K=mgh$

c) $K_{f}=\frac{1}{2}mV_{o}^{2}+mgh$

d) No

Explanation:

We have the following data:

$m$ is the mass of the projectile

$V_{o}$ is the initial speed of the projectile

$h$ is the height at which the projectile was fired

$\theta=0\°$ is the angle (it was fired horizontally)

$g$ is the acceleration due gravity

a) The Work when the applied force and the distance (height $h$) are parallel is:

$W_{g}=F_{g} hcos \theta$ (1)

Where $F_{g}=mg$ is the force exerted by gravity on the projetile (its weight)

So:

$W_{g}=mg h cos (0\°)$ (2)

$W_{g}=mgh$ (3) This is the work done by gravity

b) According to Conservation of energy principle the initial total mechanical energy $E_{o}$ is equal to the final mechanical energy  $E_{f}$:

$E_{o}=E_{f}$ (4)

Where:

$E_{o}=K_{o}+U_{o}$ (5)

Being $K_{o}=\frac{1}{2}mV_{o}^{2}$ the initial kinetic energy and $U_{o}=mgh$ the initial gravitational potential energy

$E_{f}=K_{f}+U_{f}$ (6)

Being $K_{f}=\frac{1}{2}mV_{f}^{2}$ the final kinetic energy and $U_{f}=mg(0)$ the final gravitational potential energy

So: $K_{o}+U_{o}=K_{f}+U_{f}$ (7)

$K_{f}-K_{o}=U_{o}-U_{f}$ (8)

Where $\Delta K=K_{f}-K_{o}$ is the change in kinetic energy.

Hence:

$\Delta K=U_{o}-U_{f}$

$\Delta K=mgh-mg(0)$

$\Delta K=mgh$ (9) This is the change in kinetic energy

c) Isolating $K_{f}$ from (8):

$K_{f}=U_{o}-U_{f}+K_{o}$ (10)

$K_{f}=mgh-mg(0)+\frac{1}{2}mV_{o}^{2}$

Hence:

$K_{f}=mgh+\frac{1}{2}mV_{o}^{2}$ (11) This is the final kinetic energy of the projectile

d)These results will not change if we change the angle, since these equations do not depend on the angle.