Question

When the projectile is projected at angle of projection $\theta ={45}^0,$ then range of projectile R is related to height H as:

• A. $R_{max}=2H_{max}$
• B. $R_{max}=4H_{max}$
• C. $R_{max}=H_{max}$
• D. $2R_{max}=H_{max}$

Answer 1

The correct option is B $R_{max}=4H_{max}$

Given,

$\theta ={45}^0$

The maximum height in projectile motion,

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

Substitute the value of the angle:

$H_{max}=\frac{u^{2}si{n}^2{45}^0}{2g}$

$H_{max}=\frac{u^2}{4g}$. . . . . . . .(1)

The Horizontal range of the projectile is given by,

$R=\frac{u^{2}sin2\theta }{g}$

$R_{max}=\frac{u^2\times sin{90}^0}{g}$

$R_{max}=\frac{u^2}{g}$ . . . . . . . . .(2)

From equation (1) and (2), we get

$H_{max}=\frac{R_{max}}{4}$

$4H_{max}=R_{max}$

The correct option is B.