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`        A player kicks a football obiquely at a speed of 20 meter per second so that it's range is maximum.another playerat a distance of 24 meter away in the direction of kicks starts running at that instant to catch the ball before the ball Hits the ground.the constant speed with which the 2nd player has to run is (g=10 meter per second^2)`
2 months ago

```							since range is max. range is given by [u^2 sin2(theta)]/g since range is max (theta)=45degrees.[so as sin2(theta)=1 condition for max value] and range= u^2/g =20*20/10 =40m since the other player is already 24m apart that means he has to cover a distance of (40-24)=16m time taken by ball to reach ground: t=2usin(theta)/g since theta =45degrees t=(2*20*1/root2)/10 =2*root2 speed of second player: 16/2root2 =8/root2 =4*root2
```
2 months ago
```							Maximum range of ball is R =  u^2 Sin2Q /g Q = 45 R = 40 meter Distance travelled by second man = 40-24 = 16 meter Time taken by the ball , t = 40 /uCos45 = 2sqrt(2)Speed of second man , v = distance travelled / time taken =16/2sqrt(2) = 5.65 m/sec Hope this helps Good Luck
```
2 months ago
```							Dear student,                        Maximum range of ball is                         R =  u^2 Sin2Q /g                         Q = 45                        R = 40 meter range is given by [u^2 sin2(theta)]/g since range is max (theta)=45degrees.[so as sin2(theta)=1 condition for max value] and range= u^2/g =20*20/10 =40m since the other player is already 24m apart that means he has to cover a distance of (40-24)=16m time taken by ball to reach ground: t=2usin(theta)/g since theta =45degrees t=(2*20*1/root2)/10 =2*root2 speed of second player: 16/2root2 =8/root2 =4*root2 Hope this helps you
```
14 days ago
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