Question

A projectile is projected with a kinetic energy $K$. If it has the maximum possible horizontal range, then its kinetic energy at the highest point will be

• A. $K/4$
• B. $K/2$
• C. $3K/4$
• D. $K$

Answer 1

The correct option is B $K/2$

$\text{Step 1: Maximum range of the projectile}$

Let $u$ be the initial velocity with which the projectile is projected.

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

$R_{max}=\frac{u^2}{g}$ ,when $\theta ={45}^0$

$\text{Step 2: Velocity at the highest point}$

$u_y=0$

$u_x=u\cos \theta$

Put $\theta ={45}^o$

$u_x=u\cos {45}^o$

$\therefore u_x=\frac{u}{\sqrt{2}}$ $....\left(1\right)$

$\text{Step 3: Kinetic Energy of the projectile}$

Kinetic energy at the ground, $K=\frac{1}{2}m{u}^2$

Let Kinetic Energy at the highest point be $K^{{}^{\prime }}$

$K^{\prime }=\frac{1}{2}m{u}_x^2$

Eq $\left(1\right)\Rightarrow$ $K^{\prime }=\frac{1}{2}m(\frac{u}{\sqrt{2}}{)}^2$ $\Rightarrow K^{\prime }=\frac{1}{4}m{u}^2=\frac{K}{2}$

Hence, Option(B) is correct