Question

A projectile has been thrown at an angle such that it reaches as far horizontally as it

could. The ratio of its speeds at the point of projection and at the topmost point is

______.​

Answer 1

Given : A projectile has been thrown at an angle such that it reaches as far horizontally as it  could.

To Find : The ratio of its speeds at the point of projection and at the topmost point is

Solution:

R = V₀²Sin2α / g

α = angle of projection

Sin2α is max when Sin2α  = 1

=> 2α  = 90°

=> α  = 45°

speeds at the point of projection  = V₀

Horizontal speed at at the point of projection = V₀Cosα  = V₀Cos45°

=  V₀/√2   ( and remain constant through out)

at the topmost point Vertical velocity is 0

Hence Velocity at  topmost point = Horizontal speed = V₀/√2  

The ratio of its speeds at the point of projection and at the topmost point is

= V₀ : ( V₀/√2  )

= √2  : 1

√2  : 1 is ratio of its speeds at the point of projection and at the topmost point

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Answer 2

Given:

A projectile has been thrown at an angle such that it reaches as far horizontally as it could.

To find:

The ratio of its speeds at the point of projection and at the topmost point is ?

Diagram:

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,-1){\vector(0,1){5}}\put(-1,0){\vector(1,0){8}}\qbezier(0,0)(3,5)(6,-0)\multiput(3,0)(0,0.38){7}{\qbezier(0,0)(0,0)(0,0.2)}\put(3.4,1){\bf H}\put(3,-0.7){\bf R}\put(2.8,-0.6){\vector(-1,0){2.7}}\put(3.5,-0.6){\vector(1,0){2.6}}\put(0,0){\vector(1,2){1}}\put(1.1,2.3){\bf\large u}\end{picture}$

Calculation:

Let the initial velocity of projection be "u":

Let's say max range be R :

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

For max value of R , value of sine should be be max.

$\rm \therefore \: \sin(2 \theta) = max \: value = 1$

$\rm \implies \: 2 \theta = {90}^{ \circ}$

$\rm \implies \: \theta = {45}^{ \circ}$

Now, we know that at maximum height , the net velocity is equal to the horizontal component of initial velocity of projection.

So, velocity at highest point be v:

$\rm \therefore \: v = u \sin( \theta)$

$\rm \implies\: v = u \sin( {45}^{ \circ} )$

$\rm \implies\: v = u \times \dfrac{1}{ \sqrt{2} }$

$\rm \implies\: v = \dfrac{u}{ \sqrt{2} }$

So, required ratio :

$\rm \therefore \: u : v = u : \dfrac{u}{ \sqrt{2} }$

$\rm \implies \: u : v = \sqrt{2} : 1$

So, final answer is:

$\boxed{ \bold{\: u : v = \sqrt{2} : 1}}$