Question

please answer this question
and prove it​

Answer 1

Solution :

The initial velocity of the particle is 'v'

The particle is projected at an angle θ from the horizontal

We need to find the average velocity of the particle between its point of projection and highest point of trajectory

Let us first write down the coordinates of the projectile

The coordinates of the particle at initial and highest point  are (0,0) and (R/2 ,H) respectively

pls refer the attachment first for better understanding

the r vector can be written as ,

$\displaystyle\implies \mathtt{\overrightarrow{r}=\dfrac{R}{2} \^{i} + H \^{j} }$

we know that ,

$\bigstar\boxed{\mathtt{R=\dfrac{2v^{2} sin\theta .cos \theta }{g}}}$

So,

$\bigstar\boxed{\mathtt{\dfrac{R}{2}=\dfrac{v^{2} sin\theta .cos \theta }{g}}}$

$\bigstar\boxed{\mathtt{H=\dfrac{v^{2} sin^{2}\theta }{2g}}}$

Now substituting the above values in the  equation we get ,

$\rightarrow \mathtt{\overrightarrow{r}=\dfrac{v^{2}sin\theta. cos\theta}{g} \hat{i} + \dfrac{v^{2}sin^{2} \theta }{2g} \hat{j} }$

we know that ,

$\rightarrow\mathtt{\overrightarrow{V}_{avg}=\dfrac{\overrightarrow{r}}{t}}$

The total time of flight of the particle is ,

$\bigstar\boxed{\mathtt{T=\dfrac{2vsin\theta }{g}}}$

We need to find the time taken by the particle to reach the highest point of its projection which is ,

$\bigstar\boxed{\mathtt{\dfrac{T}{2}=\dfrac{v sin\theta }{g}=t }}$

Hence ,

$\rightarrow\mathtt{\overrightarrow{V}_{avg}=\dfrac{\dfrac{v^{2}sin\theta. cos\theta}{g} \hat{i} + \dfrac{v^{2}sin^{2} \theta }{2g} \hat{j}}{\dfrac{vsin\theta}{g} }}$

$\rightarrow\mathtt{\overrightarrow{V}_{avg}=vcos\theta\:\^{i}+ \dfrac{vsin\theta }{2}\^{j}}}$

$\rightarrow\mathtt{{V}_{avg}=\sqrt{ (vcos\theta)^{2} + (\dfrac{vsin\theta }{2})^{2} }}}$

$\rightarrow\mathtt{{V}_{avg}=\sqrt{ v^{2} cos^{2}\theta + \dfrac{ v^{2} sin^{2}\theta }{4} }}}$

$\rightarrow\mathtt{{V}_{avg}=\sqrt{\dfrac{4 v^{2} cos^{2}\theta }{4}+ \dfrac{ v^{2} sin^{2}\theta }{4} }}}$

$\rightarrow\mathtt{{V}_{avg}=\dfrac{v}{2}\sqrt{ cos^{2}\theta + sin^{2} \theta}}}$

we know that ,

$\bigstar\boxed{\mathtt{sin^{2}\theta=1- cos^{2} \theta }}$

$\rightarrow\mathtt{{V}_{avg}=\dfrac{v}{2}\sqrt{ 4 cos^{2}\theta +1-cos^{2} \theta}}}$

$\rightarrow{\red{\boxed{\mathtt{{V}_{avg}=\dfrac{v}{2}\sqrt{ 3 cos^{2}\theta +1}}}}$