Question

Show that horizontal range of a body in projectile motion is maximum at 45° of projection with horizontal. Find kinetic energy at the highest point.​

Answer 1

$\large\bf{\underline{\underline{Question:-}}}$

Show that horizontal range of a body in projectile motion is maximum at 45° of projection with horizontal. Find kinetic energy at the highest point.

$\large\bf{\underline{\underline{Answer:-}}}$

Let u = velocity of projection

$\theta$ = angle of projection

$\therefore$ Horizontal range, R = $\large\frac{u²sin 2 \theta}{g}$ ______1

The range is maximum when sin 2$\theta$=1

$\implies\: sin 2 \theta = sin 90°$

$\implies\: 2 \theta = 90°$

i.e. $\theta$ = 45°

Hence, range is maximum at 45° (when there is no air resistance).

Kinetic energy at the highest point, K¹ = K cos $\theta$ , K = initial kinetic energy

$\implies$ K¹ = K cos² 45°

$\large{\boxed{K¹ = \frac{K}{2}}}$

$\red{Hope \: it \: helps}$

Answer 2

Answer:

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Kinetic energy at the highest point will be, k/2

For the range is maximum at 45°.

$\huge\boxed{\fcolorbox{black}{yellow}{HOPE \: IT \: HELPS}}$