(a) In Problem 8, using energy techniques rather than the techniques of Chapter 4, find the speed of the snowball as it reaches the ground below the cliff. What is that speed

(b) If the launch angle is changed to $\mathbf{41}\mathbf{.}\mathbf{0}\mathbf{°}$ belowthe horizontal and

(c) If the mass is changed to $\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{kg}$?

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

$h=12.5\text{m}$

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,

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

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.