Question

A football player kicked a ball from ground with initial speed of 21.3ms-1 at an angle of 45. Another player 55m away in the direction of kick starts to meet the ball at that instant. what must be his average speed, if he meet the ball just before hit to the ground.​

Answer 1

Given:

A football player kicked a ball from ground with initial speed of 21.3ms-1 at an angle of 45. Another player 55m away in the direction of kick starts to meet the ball at that instant.

To find:

His average speed, if he meet the ball just before hit to the ground.

Calculation:

This is an example of Ground to Ground projectile.

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

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

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

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

$= > \: range = \dfrac{ {(21.3)}^{2} }{10}$

$= > \: range = 45.36 \: m$

Let time taken to complete trajectory be t:

$\therefore \: t = \dfrac{2u \sin( \theta) }{g}$

$= > \: t = \dfrac{2(21.3) \sin( {45}^{ \circ} ) }{10}$

$= > \: t = 3.012 \: sec$

So, distance to be travelled in this time by the 2nd player = 55 - 45.36 = 9.64 m

$\therefore \: avg. \: v = \dfrac{9.64}{3.012}$

$= > \: avg. \: v = 3.2 \: m {s}^{ - 1}$

So, his average speed will be 3.2 m/s.