Question

An object is thrown at a projection angle of 45° with a velocity of v0. Find the distance between the point at which the object is thrown and the point with the maximum height it assumes.​

Answer 1

Formulae to be used :-

• $\sf R = \dfrac{u^2\;\;sin(2\theta)}{g}$
• $\sf H = \dfrac{u^2\;\;sin^2(\theta)}{2g}$

Answer:-

We have to find the distance between the point of projection (A) and the point at which the maximum height is achieved (B).

So, we have to find the length of AB.

For that, first we have to find the length of AC and BC.

Finding AC:-

We can clearly see that, AC is half the horizontal range of the projectile.

$\sf AC = \dfrac{R}{2}$

$\longrightarrow \sf AC = \dfrac{\bigg(\dfrac{u^2\;sin(2\theta)}{g}\bigg)}{2}$

$\longrightarrow \sf AC = \dfrac{u^2\;\;sin(2\theta)}{2g}$

Putting $\sf u = v_0$ and $\sf \theta = 45^o$

$\longrightarrow \sf AC = \dfrac{v_0^2\;\; sin(2\times 45^o)}{2g}$

$\longrightarrow \sf AC = \dfrac{v_0^2}{2g}\quad \quad \dots (1)$

Finding BC:-

We can clearly see that, the length of BC is equal to the maximum height attained by the projectile.

$\sf BC = \dfrac{u^2\;\;sin^2\theta}{2g}$

Putting $\sf u = v_0$ and $\sf \theta = 45^o,$

$\longrightarrow \sf BC = \dfrac{v_0^2 \;\; sin^2(45^o)}{2g}$

$\longrightarrow \sf BC = \dfrac{\bigg(v_0^2 \times \dfrac{1}{2}\bigg)}{2g}$

$\longrightarrow \sf BC = \dfrac{v_0^2}{4g} \quad \quad \dots (2)$

Now that we have the length of AC and BC, we can find the length of AB.

Finding AB:-

In $\sf \triangle ABC$ right angled at C,

$\sf AB^2 = BC^2 + AC^2$

$\small \longrightarrow \sf AB^2 = \bigg(\dfrac{v_0^2}{4g}\bigg)^2 + \bigg(\dfrac{v_0^2}{2g}\bigg)^2 \quad [From\:(2)\:and\:(1)]$

$\longrightarrow \sf AB^2 = \dfrac{v_0^4}{16g^2} + \dfrac{v_0^4}{4g^2}$

$\longrightarrow \sf AB^2 = \dfrac{v_0^4}{4g^2}\bigg( \dfrac{1}{4} + 1\bigg)$

$\longrightarrow \sf AB = \sqrt{\dfrac{v_0^4}{4g^2}\times \dfrac{5}{4}}$

$\longrightarrow \sf AB = \dfrac{v_0^2}{2g}\times \dfrac{1}{2} \times \sqrt{5}$

$\longrightarrow \sf \underline{\underline{\sf{\green{ AB = \dfrac{\sqrt{5}v_0^2}{4g} }}}}$

Answer 2

Given:

An object is thrown at a projection angle of 45° with a velocity of v_(0).

To find:

The distance between the point at which the object is thrown and the point with the maximum height it assumes.

Calculation:

Let the max height is obtained at half range:

$\rm \therefore \: x = \dfrac{r}{2}$

$\rm \implies \: x = \dfrac{1}{2} \times \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

$\rm \implies\: x = \dfrac{ {( v_{0}) }^{2} \sin(2 \theta) }{2g}$

$\rm \implies\: x = \dfrac{ {( v_{0}) }^{2} \sin(2 \times {45}^{ \circ} ) }{2g}$

$\rm \implies\: x = \dfrac{ {( v_{0}) }^{2} \sin({90}^{ \circ} ) }{2g}$

$\rm \implies\: x = \dfrac{ {( v_{0}) }^{2} }{2g}$

Now , let max height be h :

$\rm\therefore \: h = \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{2g}$

$\rm\implies \: h = \dfrac{ {( v_{0})}^{2} { \sin}^{2}( {45}^{ \circ} ) }{2g}$

$\rm\implies \: h = \dfrac{ {( v_{0})}^{2} \times \frac{1}{2} }{2g}$

$\rm\implies \: h = \dfrac{ {( v_{0})}^{2} }{4g}$

Now, the required distance is :

$\rm \therefore \: d = \sqrt{ {h}^{2} + {x}^{2} }$

$\rm \implies \: d = \sqrt{ { \bigg \{\dfrac{ { (v_{0})}^{2} }{4g} \bigg \} }^{2} + { \bigg \{\dfrac{ { (v_{0})}^{2} }{2g} \bigg \} }^{2}}$

$\rm \implies \: d = \dfrac{ { (v_{0})}^{2} }{g} \sqrt{ \dfrac{1}{16} + \dfrac{1}{4} }$

$\rm \implies \: d = \dfrac{ { (v_{0})}^{2} }{g} \sqrt{ \dfrac{1 + 4}{16} }$

$\rm \implies \: d = \dfrac{ { (v_{0})}^{2} }{g} \sqrt{ \dfrac{5}{16} }$

$\rm \implies \: d = \dfrac{ \sqrt{5} { (v_{0})}^{2} }{4g}$

So, the required distance is:

$\boxed{ \bf{ \red{\: d = \dfrac{ \sqrt{5} { (v_{0})}^{2} }{4g} }}}$