91Ó°ÊÓ

$v_i=14.\text{0m/s}$

$h=12.5\text{m}$

The problem is based on the law of conservation of energy which states that the total energy of an isolated system remains constant.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula:

$\text{Initial}  \text{total}  \text{energy}  \text{=}  \text{Final}  \text{total}  \text{energy}$

According to the law of conservation of energy,

$K{E}_i+P{E}_i=K{E}_f+P{E}_f$

$\frac{1}{2}m{v}_i^2+mgh=\frac{1}{2}m{v}_f^2+0$

Hence,

$\begin{array}{rcl}{v}_f& =& \sqrt{v_i^2+2gh}\\ & =& \sqrt{{14}^2+\left(2×9.8×12.5\right)}\\ & =& 21\text{m/s}\end{array}$

Hence, the speed of the snowball as it reaches the ground below the cliff is 21 m/s.

Here, final velocity does not change.

Hence,

$v_f=21\mathrm{m}/\mathrm{s}$

Hence,the speed if the launch angle is changed to $41.0^{\text{o}}$below the horizontal is 21 m/s.

Velocity found in part (a) doesn’t depend on mass. So, it would bethe same.

$v_f=21\mathrm{m}/\mathrm{s}$

Hence, the speed if the mass is changed to 2.50 kg is 21 m/s.

Therefore, the unknown speed can be found by using the law of conservation of energy.