You are standing at the top of a 50 mcliff. You throw a rock in the horizontal direction with speed 10 m/s. If you neglect air resistance, where would you predict it would hit on the flat plain below? Is your prediction too large or too small as a result of neglecting air resistance?

height of the hill is h=50.0m, horizontal velocity ${\mathrm{u}}_{\mathrm{x}}=10\mathrm{m}/\mathrm{s}$, vertical velocity , horizontal velocity ${\mathrm{u}}_{\mathrm{x}}=10\mathrm{m}/\mathrm{s}$, acceleration along the horizontal direction${\mathrm{a}}_{\mathrm{x}}=0$ and acceleration along the vertical direction ${\mathrm{a}}_{\mathrm{y}}=\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2$

Internal energy, which emerges from the molecular state of motion of matter is an energy form that exists in every system.Internal energy is represented by the letter and the Joule is the unit of measurement.

Let the time be

Apply the equations of motion along the vertical direction.

$$\begin{gathered} h=u_{y}t+\frac{1}{2}{a}_y{t}^2 \\ h=0+\frac{1}{2}g{t}^2 \\ t=\sqrt{\frac{2h}{g}} \end{gathered}$$

Substitute$g=9.8m/s^2$ and$h=50$ into the obtained formula.

$$\begin{gathered} \mathrm{t}=\sqrt{\frac{2\left(50.0\mathrm{m}\right)}{9.8\mathrm{m}/{\mathrm{s}}^2}} \\ =3.19\mathrm{s} \end{gathered}$$

Therefore, the time of decent is 3.19 s .

Let the horizontal distance travelled by the rock bex.

Apply the equation of motion along the horizontal direction.

$$\begin{gathered} x=u_{x}t+\frac{1}{2}{a}_x{t}^2 \\ =u_{x}t+0 \\ =u_{x}t \end{gathered}$$

Substitute ${\mathrm{u}}_{\mathrm{x}}=10\mathrm{m}/\mathrm{s}\mathrm{and}\mathrm{t}=3.19\mathrm{s}$into the obtained formula.

$$\begin{gathered} x=\left(10m/s\right)\left(3.19s\right) \\ =32.0m \end{gathered}$$

As a result, the rock's horizontal distance travelled from the top is $x=32.0m$and the predicted distance is said to be long in the absence of air resistance.