91Ó°ÊÓ

The given data is listed below as-

• The height of the rock above the ground is $h=15 m$
• The final speed of rock is,$V_2=25 m/s$

When forces act on a particle it undergoes displacement and the particle’s kinetic energy changes by an amount equal to the total work done on the particle by all the forces. Therefore, work done on the particle is given by-

${\mathbf{W}}_{\mathbf{total}}\mathbf{=}{\mathbf{K}}_{\mathbf{2}}\mathbf{-}{\mathbf{K}}_{\mathbf{1}}\mathbf{=}\mathbf{∆}\mathbf{K}$

The work-energy theorem can be applied to all the bodies that can be treated as particles.

(a)

Use the work-energy theorem,

$$\begin{gathered} W_{net}=K_2-K_1 \\ Fh=\frac{1}{2}m{V}_2^2-\frac{1}{2}m{V}_1^2 \end{gathered}$$

Now, the work done is negative since gravitational force is opposite to the direction of displacement.

$-mgh=\frac{1}{2}m{V}_2^2-\frac{1}{2}m{V}_1^2$

Divide the above equation by m to obtain the equation as below:

$-gh=\frac{1}{2}{V}_2^2-\frac{1}{2}{V}_1^2$

Here, m is the gravitational constant, h is height of the rock above ground, V₁is the initial velocity of rock, and V₂is the final velocity of the rock.

For, $g=9.8m{s}^{-2},h=15m,\mathrm{and}{V}_2=25m/s$

$-gh=\frac{1}{2}{V}_2^2-\frac{1}{2}{V}_1^2$will be

$\left(9.8{\mathrm{ms}}^{-2}\right)=\frac{1}{2}{\left(25{\mathrm{ms}}^{-1}\right)}^2-\frac{1}{2}{\mathrm{V}}_1^2$

Therefore, solve the above equation and obtain the value of V₁ as below:

$V_1=30.0m/s$

Thus, the speed of rock as it just left the ground is 30.3m/s.

(b)

The velocity at maximum height will be 0.

$-mg{h}_{max}=\frac{1}{2}m{V}_2^2-\frac{1}{2}m{V}_1^2$

Divide the above equation by m to obtain the equation as below:

$-g{h}_{max}=\frac{1}{2}m{V}_2^2-\frac{1}{2}m{V}_1^2$

Here, m is the gravitational constant, V₁is the initial velocity of rock, and V₂is the final velocity of the rock.

For, $g=9.8m{s}^{-2},V_2=0,\mathrm{and}{V}_2=25m/s$

$-g{h}_{max}=\frac{1}{2}{V}_2^2-\frac{1}{2}{V}_1^2$ will be

$$\begin{gathered} -\left(9.8m{s}^{-2}\right)h_{max}=0-\frac{1}{2}{\left(30.3m{s}^{-1}\right)}^2 \\ {h}_{max}=46.8m \end{gathered}$$

Therefore, the maximum height is 46.8 m