X player 64 @ speed of 20 m per second so that its horizontal range is maximum another play 24 m away in the direction of kick start earning in the same direction at the same instant of it if he has to catch the ball just before it reaches the ground should run with velocity is equal to
Answers
we know that the range is MAXIMUM WHEN THE ANGLE BETWEEN PROJECTILE AND THE HORIZONTAL IS 45 *. SO , horizontal range of the ball = u ^2 sin2x / g i.e. equal to (20 x 20 x sin 2(45) ) /2 = 200 m . Now we know that ball will land on ground 200 m away from player . Let us now calculate the time it takes to reach that position , we know the formula of time of flight = 2 u sin x/g
putting values you will get 2 root 2 secs i.e equal to 2.8 seconds . now we see that ball takes 2.8 seconds to land 200 m away from player 1 . player two is already 24 m away from player 1 . so the ball will land ( 200-24)m away from him . it will be equal to 176 m, . now he has to catch the ball before landing . he has to cover 176m in 2.8 seconds . we know the relation between speed , dist. and time . i.e. s=d/t .... solving this you will get s= 176/2.8 i.e. 6.2 m / sec.
There might be a calculation error maybe but you have to follow the same procedure .