# 11. Physics | Kinematics | Ball Caught by a Fielder in a Cricket Match | by Ashish Arora

in this illustration, we’ll see a situation

when a ball is caught by a fielder in a cricket match. here we are given that a batsman hits

a shot and ball leaves his bat at 20 meter per second at an angle 45 degree. and a fielder

located 25 meter away. starts running along the line of balls motion. and it is asking

what must be his speed so that he will be able to catch the ball. and we can neglect

the height of the fielder. here in situation if we draw the figure, you can see. on ground

from a point. a short is fired at an angle, 45 degree with the speed 20 meters per second.

and it follows a projectile path. and at a distance 25 meter away from the batsman. a

fielder is located. the height of fielder we are given to neglect. he starts running

in such a way that he has to catch the ball, so we are required to find the speed of this

fielder so that, the ball will be caught. here in this situation first we can find out

the range of ball, after. the short. in this situation the range we can directly write

as theta is 45 degree it is maximum range which is u square by g. so we can substitute

the values, the speed is 20 square by g. so this is 40 meter. that means if this distance

is 25 meter the remaining distance where the ball will, lying down ground is 15 meter here,

the figure is not in proper proportion. here we can write also that not in proportion.

this i have drawn, that remaining distance is 15 meter. so here, i can also calculate

the time of flight, of the ball. which can be given as 2 u sine theta by g. the value

will be 2, speed is 20 and sine 45 is 1 by root 2. divided by g. so this can be written

as 2 root 2 seconds. so in this much of time, the fielder has to travel a distance 15 meter

so, we can directly write the velocity required. for fielder. to catch the ball. can be written

as, v, which is equal to 15 divided by, 2 root 2. so, this will be, numerically if you

calculate it is 5 point 3 meters per second that is the result of this problem.

Easy question who like in this tough sires

Sir I did as time taken by ball to travel 25metre= 25√2/20=5/2√2

Now time left for player to catch the ball=2√2-5/2√2=3/2√2.

So required velocity=(15*2√2)/3=10√2m/sec and why this is not right