Chapter 1.7, Problem 22P

A)

Interpretation Introduction

Interpretation:

The velocity of dropped object when hits the ground

Concept introduction:

Law of conservation of energy:

  The energy can neither be created nor be destroyed but it changes from one form to another. Accordingly, the energy entering into the open system will be equivalent to the energy leaving out of the same system, thereby conserving energy.

$\begin{array}{l}\Delta \text{P}\text{.E=}\Delta \text{K}\text{.E}\\ \Delta \text{P}\text{.E}+\Delta \text{K}\text{.E}=0\\ Mg\Delta z+\frac{M\Delta v^2}{2}=0\\ \frac{\left(v_2^2-v_1^2\right)}{2}=-g\left(z_2-z_1\right)\\ v_2^2=\sqrt{v_1^2+\left[-2g\left(z_2-z_1\right)\right]}\end{array}$

Here, gravity is g, mass is M, velocity is v, initial height is $z_1$, final height is $z_2$, and difference in height is $\Delta z$.

B)

Interpretation Introduction

Interpretation:

The velocity of thrown down object when hits the ground

Concept introduction:

Law of conservation of energy:

  The energy can neither be created nor be destroyed but it changes from one form to another. Accordingly, the energy entering into the open system will be equivalent to the energy leaving out of the same system, thereby conserving energy.

The law of energy conservation for the thrown ball

$\begin{array}{l}\Delta \text{P}\text{.E}+\Delta \text{K}\text{.E}=0\\ Mg\Delta z+\frac{M\Delta v^2}{2}=0\\ \frac{\left(v_2^2-v_1^2\right)}{2}=-g\left(z_2-z_1\right)\\ v_2=\sqrt{2g\left(z_1-z_2\right)+v_1^2}\end{array}$

C)

Interpretation Introduction

Interpretation:

Examples for the assumption made in part A and B have to be determined.

Concept introduction:

Air resistance is depending on the ratio of surface area to the mass, if the mass is less and the surface area is more means air resistance is more.