A $1.50\text{kg}$ snowball is fired from a cliff$12.5\text{m}$ high. The snowball’s initial velocity is$14.0\text{m/s}$ , directed $41.0\text{°}$above the horizontal. (a) How much work is done on the snowball by the gravitational force during its flight to the flat ground below the cliff? (b) What is the change in the gravitational potential energy of the snowball-Earth system during the flight? (c) If that gravitational potential energy is taken to be zero at the height of the cliff, what is its value when the snowball reaches the ground?

i) Mass of snowball$m=1.50\text{\hspace{0.17em}kg}$

ii) Height through which the snowball drops$h=12.5\text{\hspace{0.17em}m}$

iii) Gravitational acceleration $g=9.8{\text{m/s}}^{\text{2}}$

iv) The angle ${\theta }_0={41.0}^{\text{o}}$

v) Initial velocity$v_0=14.0\text{m/s}$

Find the work done due to gravity using the formula in terms of gravitational force and height of the object.

i) Gravitational work on ball is given by formula

ii) Change in potential energy is given by

iii) Potential energy

Theforce is vertically downward and has magnitude mg, where mis the mass of the snowball.

$W_g=mgh$

$\Rightarrow W_g=1.50\times 9.80\times 12.5$

$\Rightarrow W_g=184\text{\hspace{0.17em}J}$

The force of gravity is conservative, so the change in the potential energy of the snowball Earth system is negative of the work that it does

$\Delta U=-W_g=-mgh$

$\Rightarrow \Delta U=-mgh$

$\Rightarrow \Delta U=-1.50\times 9.80\times 12.5$

$\Rightarrow \Delta U=-184\text{\hspace{0.17em}J}$

The potential energy, when it reaches the ground, is less than the potential energy when it is fired so, the potential energy when the snowball hits the ground is

$U=-mgh$

$\Rightarrow U=-1.50\times 9.80\times 12.5$

$\Rightarrow U=-184\text{\hspace{0.17em}J}$