Q62P - University Physics with Modern Physics

Question

A rock is thrown with a velocity ${\mathbf{V}}_{\mathbf{0}}$, at an angle of ${\mathbf{\alpha }}_{\mathbf{0}}$ from the horizontal, from the roof of a building of height h. Ignore air resistance. Calculate the speed of the rock just before it strikes the ground, and show that this speed is independent of ${\mathit{\alpha }}_{\mathbf{0}}$.

Step-by-Step Solution

Answer

The speed $v=\sqrt{{v_0}^2+2gh}$ is independent by ${\alpha }_0$.

Step 1: Introduction:

According to the law of conservation of energy, which is found in physics, the total energy of an isolated system is constant and is said to be conserved over time.

Step 2: Consider the known data as below.

Angle is ${\alpha }_0$.

The mass of the rock is   $m$.

Acceleration due to gravity is g.  

Initial velocity is $v_0$.

Final velocity of the rock is v. 

Vertical distance from the ground is h.

Step 3: Explanation:

The initial kinetic energy is,

${\mathrm{KE}}_1=\frac{1}{2}{\mathrm{mv}}_0^2$ 

 

The initial potential energy is,

${\mathrm{PE}}_1=\mathrm{mgh}$ 

 

The final kinetic energy is,

${\mathrm{KE}}_2=\frac{1}{2}{\mathrm{mv}}^2$ 

 

The final potential energy is,

${\mathrm{PE}}_2=0$ 

 

Since the height is zero.

 

According to conservation of energy,

$$\begin{gathered} {\mathrm{KE}}_1+{\mathrm{PE}}_1={\mathrm{KE}}_2+{\mathrm{PE}}_2 \\ \frac{1}{2}{\mathrm{mv}}_0^2+\mathrm{mgh}=\frac{1}{2}{\mathrm{mv}}^2+0 \\ {\mathrm{v}}_0^2+2\mathrm{gh}={\mathrm{v}}^2 \\ \mathrm{v}=\sqrt{{\mathrm{v}}_0^2+2\mathrm{gh}} \end{gathered}$$ 

And it can be seen that the velocity is independent of the angle ${\alpha }_0$.