Question

An object is projected at an angle of 45 degree with the horizontal. The horizontal range and the maximum height reached will be in ratio.

Answer 1

Range = u²sin(2θ) / g
= u²sin(2 × 45) / g
= u²sin(90) / g
= u²/g

Maximum height = u²sin²θ / (2g)
= u²sin²(45) / (2g)
= (u² × 1/2) / (2g)
= u² / (4g)

Ratio = [u²/g] : [u² / (4g)] = 1 : 1/4 = 4 : 1

Answer 2

Answer:

The ratio of horizontal range (R) and the maximum height (H) reached will be 4 : 1.

Explanation:

Given, the angle of the projectile with horizontal θ= 45°

The range of the projected body is given by:

$R=\frac{u^{2}Sin2\theta}{g}$                                                                       .................(1)

where u is the initial velocity of the projectile,

g is the acceleration due to gravity.

Put the value of angle θ= 45° in equation (1);

$R=\frac{u^{2}sin90^{o}}{g}$

$R=\frac{u^{2}}{g}$

Height of the projectile (projected body) is given by:

$H=\frac{u^{2}sin^2\theta}{2g}$

$H=\frac{u^{2}(sin45^{o})^{2}}{2g}$

$H=\frac{u^{2}(\frac{1}{\sqrt{2} })^{2} }{2g}$

$H=\frac{u^{2}}{4g}$

The ratio of horizontal range and the maximum height reached will be:

$\frac{R}{H}=\frac{{u^{2}}/{g} }{{u^{2}}/{4g} }$

$\frac{R}{H} =\frac{4}{1}$

Therefore, the ratio of the range (R) and height (H) will be 4 : 1.