91Ó°ÊÓ

When gravity first exerts force on an item, its initial velocity indicates how fast it travels. The final velocity, on the other hand, is a vector number that measures a moving body's speed and direction after it has reached its maximum acceleration.

The velocity can be said to be V_i. The velocity is making some angle with the x axis.

$V_{iy}=V_i\sin \mathrm{θ}$

The displacement will be$Y=0$meters.

The time is not given.

The gravitational acceleration is $-9.8m/s^2$$-9.8m/s^2$.

The initial velocity and the final velocity and the average velocity will be same.

$$\begin{gathered} \stackrel{¯}{V}=\frac{X}{t} \\ t=\frac{X}{V_i\cos \mathrm{θ}}..........1 \end{gathered}$$

Now let’s calculate the displacement.

$$\begin{gathered} ∆Y=V_{iy}+\frac{1}{2}{g}_y{t}^2 \\ 0=V_i\sin \mathrm{θ}×t+\frac{1}{2}(-g){\left(t\right)}^2\frac{1}{2}\left(g\right){\left(t\right)}^2=V_i\sin \mathrm{θ}×t \\ \frac{1}{2}\left(g\right)\left(t\right)=V_i\sin \mathrm{θ} \\ t=\frac{2V_i\sin \mathrm{θ}}{g} \end{gathered}$$

The time when displacement is zero is in the above equation.

The average velocity will be equal to the x component of initial velocity.

$\stackrel{¯}{V}=V_{ix}=V_i\cos \mathrm{θ}$

Also

$$\begin{gathered} \stackrel{¯}{V}=\frac{∆X}{t} \\ ∆X=\stackrel{¯}{V}×t \end{gathered}$$

Putting the values of v and t into the above equation, we have

$$\begin{gathered} \stackrel{¯}{V}=\frac{∆X}{t} \\ ∆X=V_i\cos \mathrm{θ}×\left(\frac{2V_i\sin \mathrm{θ}}{g}\right) \\ ∆X=\left(\frac{2{V_i}^2\sin \mathrm{θ}\cos \mathrm{θ}}{g}\right) \\ ∆X=\left(\frac{{V_i}^2\left[2\sin \mathrm{θ}\cos \mathrm{θ}\right]}{g}\right) \\ ∆X=\left(\frac{{V_i}^2\left[\sin 2\mathrm{θ}\right]}{g}\right) \\ ∆X=\left(\frac{{V_i}^2\left[\sin 2\mathrm{θ}\right]}{g}\right) \\ R=\left(\frac{{V_i}^2\sin 2\mathrm{θ}}{g}\right) \end{gathered}$$

Hence the range of the projectile motion is derived as above.