Question

A body is projected with a velocity of 30m/s at an angle of 30°from its horizontal. what is the maximum horizontal range of the projectile.​

Answer 1

Answer:

The projected velocity of the body u = 30 m/s.

The angle of inclination of the velocity with vertical is 30 degree.

Hence, the horizontal component of the velocity u_{x}=usin30u

x

$</p><p> =usin30</p><p></p><p> = \frac{u}{2} </p><p>2</p><p>u$

$= \frac{30}{2}\ m/s </p><p>2</p><p>30</p><p> </p><p> m/s$

= 15 m/s

The vertical component of the velocity u_{y}\ =\ ucos30u

y

= ucos30

=30\times

$\frac{\sqrt{3}} {2}\ m/s=30× </p><p>2</p><p>3</p><p> </p><p> </p><p> </p><p> m/s</p><p></p><p> =15\sqrt{3}\ m/s=15 </p><p>3</p><p> </p><p> m/s$

The maximum height is calculated as -

$h_{max} =\frac{u^2(cos\theta)^2}{2g}h </p><p>max</p><p> </p><p> = </p><p>2g</p><p>u </p><p>2</p><p> (cosθ) </p><p>2</p><p>$

$=(30)^2(\sqrt{3}/2)^2\times\frac{1}{2\times10}\ m=(30) </p><p>2</p><p> ( </p><p>3</p><p> </p><p> /2) </p><p>2</p><p> × </p><p>2×10</p><p>1</p><p> </p><p> m$

=33.75\ m=33.75 m

The total time of flight is calculated as -

$</p><p> T=2\frac{ucos\theta}{g}T=2 </p><p>g</p><p>ucosθ$

$=2\frac{30\times }{}$

$</p><p></p><p> =2\frac{30\times \sqrt{3}/2} {10}\ s=2 </p><p>10</p><p>30× </p><p>3</p><p> </p><p> /2</p><p> </p><p> s</p><p></p><p> =3\sqrt{3}\ s=3 </p><p>3</p><p> </p><p> s$

The horizontal range is calculated as -

$R = \frac{u^2sin2\theta}{g}R= </p><p>g</p><p>u </p><p>2</p><p> sin2θ$

$\frac{30^2\times \sqrt3}{/2}$

${10}\ m= </p><p>10</p><p>30 </p><p>2</p><p> × </p><p>3</p><p> </p><p> /2</p><p> </p><p> m</p><p></p><p> =45\sqrt{3}\ m=45 </p><p>3</p><p> </p><p> m$