Question

A particle is thrown with initial speed u at angle ‘θ’, with the horizontal, under gravity. The relation between maximum height ‘h’ and range upon level ground ‘R’ is

Answer 1

Given:

A particle is thrown with initial speed u at angle ‘θ’, with the horizontal, under gravity.

To find:

Relation between R and H?

Calculation:

• Max height (H) is given as :

$H = \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{2g}$

• Range (R) is given as :

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

Now, dividing the equations :

$\implies \: \dfrac{H}{R} = \dfrac{ { \sin}^{2} ( \theta)}{ 2\sin(2 \theta) }$

$\implies \: \dfrac{H}{R} = \dfrac{ { \sin}^{2} ( \theta)}{ 4\sin( \theta) \cos( \theta) }$

$\implies \: \dfrac{H}{R} = \dfrac{ \tan ( \theta)}{ 4 }$

$\boxed{ \implies \: H = \dfrac{ R\tan ( \theta)}{ 4 } }$

So, this is the relation between H and R.

Answer 2

Given :-

A particle is thrown with velocity u and with angle θ.

As we know that,

R = u²Sin2θ/g

R = u² × 2Sinθ•Cosθ/g

R = 2u²Sinθ•Cosθ/g -------(1)

H = u²Sin²θ/2g -----(2)

On dividing (2) with (1).,

H/R = u²Sin²θ/2g × g/2u²Sinθ•Cosθ

H/R = Sinθ/4Cosθ

H = R tanθ/4

Hence,

The relation between max. height and Range = H = R tanθ/4