Question

The range of a particle when launched at an angle $5^o$ with horizontal with certain velocity is $1.5km$. What is the range of the projectile when launched at an angle of ${45}^o$ to the horizontal with the same velocity?

• A. equal to $1.5km$
• B. greater than $1.5km$
• C. less than $1.5km$
• D. range at ${45}^o$ is $1.2km$

Answer 1

The correct option is B greater than $1.5km$

Let the projectile speed be $u$.

Case 1 :

$\theta =5^o$

Range $R=1.5km$

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

$\therefore$ $1.5=\frac{u^2\times sin\left(10\right)}{g}$ .......(1)

Case 2 :

$\theta ={45}^o$

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

$\therefore$ $R=\frac{u^2\times sin\left(90\right)}{g}$ .......(2)

Dividing (1) and (2), we get $\frac{R}{1.5}=\frac{sin90}{sin10}=\frac{1}{\sin 10}$

$⟹R=\frac{1.5}{\sin 10}>1.5km$

Thus option B is correct.