Question

Derive $\mathit{R}\mathbf{=}\frac{{{\mathbf{V}}_{\mathbf{o}}}^{\mathbf{2}}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{2}{\mathbf{\theta }}_{\mathbf{o}}}{\mathbf{g}}$for the range of a projectile on level ground by finding the time at which becomes zero and substituting this value of into the expression for${\mathit{X}}_{\mathbf{1}\mathbf{}}{\mathit{X}}_{\mathbf{0}}$ , noting that $\mathit{R}\mathbf{=}{\mathit{X}}_{\mathbf{1}}\mathbf{}{\mathit{X}}_{\mathbf{0}}$.

Short answer

The range of the projectile motion is derived as required.

Step-by-step solution

Definition of initial velocity

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.

Stating given data

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

$V_{iy}=V_i\sin \theta$

The displacement will be$Y=0$meters.

The time is not given.

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

Deriving parabolic trajectory of displacement

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

$$\begin{gathered} \overline{V}=\frac{X}{t} \\ t=\frac{X}{V_i\cos \theta }..........1 \end{gathered}$$

Now let’s calculate the displacement.

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

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

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

$\overline{V}=V_{ix}=V_i\cos \theta$

Also

$$\begin{gathered} \overline{V}=\frac{∆X}{t} \\ ∆X=\overline{V}\times t \end{gathered}$$

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

$$\begin{gathered} \overline{V}=\frac{∆X}{t} \\ ∆X=V_i\cos \theta \times \left(\frac{2V_i\sin \theta }{g}\right) \\ ∆X=\left(\frac{2{V_i}^2\sin \theta \cos \theta }{g}\right) \\ ∆X=\left(\frac{{V_i}^2\left[2\sin \theta \cos \theta \right]}{g}\right) \\ ∆X=\left(\frac{{V_i}^2\left[\sin 2\theta \right]}{g}\right) \\ ∆X=\left(\frac{{V_i}^2\left[\sin 2\theta \right]}{g}\right) \\ R=\left(\frac{{V_i}^2\sin 2\theta }{g}\right) \end{gathered}$$

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