SolutionverifiedVerified by TopprCorrect option is A)
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =g
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sec
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 +
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 )
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 )
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h=
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2g
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2g n
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2g n 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2g n 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2g n 2 ⇒
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2g n 2 ⇒ hαn
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2g n 2 ⇒ hαn 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2g n 2 ⇒ hαn 2
Distance Covered represents the total distance covered by a defender from the time the bat makes contact with the ball until the moment he fields it. This metric only takes into account the route actually traveled by the fielder -- NOT the direct route from the fielder's starting position to the ball.
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Answer:
✏️Solution
Solutionverified
SolutionverifiedVerified by Toppr
SolutionverifiedVerified by TopprCorrect option is A)
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =g
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sec
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 +
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 )
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 )
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h=
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2g
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2g n
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2g n 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2g n 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2g n 2 ⇒
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2g n 2 ⇒ hαn
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2g n 2 ⇒ hαn 2
SolutionverifiedVerified by TopprCorrect option is A)Let h be height of tower acceleration =ginitial velocity u=0time =n sech=ut+ 2at 2 h=0×n 2 + 2g(n 2 ) h= 2g n 2 ⇒ hαn 2
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