Question

A Player kicks a ball at an angle of 45 to the horizontal with initial Speed of
20m/s. A second player 60m away in the direction of kick starti immediaty to catch the ball.
The Speed With which the second player should run so that he can catch the ball
Just before it hit the ground (g: 10m/s-2)​

Answer 1

Answer :

Angle of projection = 45°

Initial speed = 20m/s

Second player is standing at a distance of 60m away from the projection point.

■ First of all we need to find range and time of flight of the projectile body.

❇ Range (R) :

$:\implies\sf\:R=\dfrac{u^2sin2\theta}{g}$

$:\implies\sf\:R=\dfrac{(20)^2sin2(45^{\circ})}{10}$

$:\implies\sf\:R=\dfrac{400sin90^{\circ}}{10}$

$:\implies\sf\:R=40(1)$

$:\implies\bf\:R=40\:m$

❇ Time of flight (T) :

$:\implies\sf\:T=\dfrac{2usin\theta}{g}$

$:\implies\sf\:T=\dfrac{2(20)sin45^{\circ}}{10}$

$:\implies\sf\:T=\dfrac{40}{10}\times\dfrac{1}{\sqrt2}$

$:\implies\sf\:T=\dfrac{4}{\sqrt2}$

$:\implies\sf\:T=2\sqrt2\:s$

In order to catch the ball, player has to cover the distance equal to 60 - 40 = 20m in 2√2s.

➨ speed = distance/time

➨ speed = 20/2√2

➨ speed = 10/√2

➨ speed = 5√2 m/s

Answer 2

Answer:

5√2 m/s

Explanation:

It is a limiting case i.e the player the ball will be just hitting the ground when player 2 will reach it

Use the formula of time of flight of projectile

T = (2u Sin∅)/g

EXTRA STUFF →

Horizontal range is maximum when angle of projection is 45°

Time of flight = {2u( in y direction)}/g