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Grade 11Mechanics

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)

Profile image of PRIYANK DANAVLE
6 Years agoGrade 11
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3 Answers

Profile image of Arun
6 Years 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
 
Profile image of Vikas TU
6 Years 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 
Profile image of aswanth nayak
6 Years 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