Question

A man can throw a ball 50m vertically upwards.Find the horizontal distance

Answer 1

u² sin²θ / 2g. Maximum height is obtained when θ = 90°, i.e., when the body is projected vertically upwards. That's why the answer is H = 100 / 2 = 50 m

Answer 2

Given:

A man can throw a ball 50m vertically upwards.

To find:

The maximum horizantal distance travelled by the ball if thrown with same Velocity.

Calculation:

In vertical trajectory:

$\therefore \: {v}^{2} = {u}^{2} - 2gh$

$= > \: {(0)}^{2} = {u}^{2} - (2 \times 10 \times 50)$

$= > \: {u}^{2} = 1000 \: \: \: \: ......(1)$

In the Projectile Trajectory , the maximum range is obtained when angle of Projection is 45°.

So, max range:

$\therefore \: R_{max} = \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

$= > \: R_{max} = \dfrac{ {u}^{2} \sin(2 \times {45}^{ \circ} ) }{g}$

$= > \: R_{max} = \dfrac{ {u}^{2} \sin( {90}^{ \circ} ) }{g}$

$= > \: R_{max} = \dfrac{ {u}^{2} }{g}$

$= > \: R_{max} = \dfrac{ 1000 }{g}$

$= > \: R_{max} = \dfrac{ 1000 }{10}$

$= > \: R_{max} = 100 \: m$

So, final answer is:

$\boxed{ \bf{ \red{\: R_{max} = 100 \: m}}}$