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A cricketer hits a ball with a velocity 25 m/s at 60° above the horizontal. How far above the ground it passed over a fielder 50 m from the bat (assume the ball is struck very close to the ground)
(a) 8.2m
(b) 9.0m
(c) 11.6m
(d) 12.7m
horizontal component of velocity = Vx = Ucos@
Ux = 25cos60=25/2 m/s
Uy = 25sin60 = 25(3)1/2 /2
it covers a distance of 50 m after time t
since accleration along horizontal direction is 0 so
V = d/t
25/2 = 50/t
t = 4sec
now after 4 sec let the height attained by particle is H then
H = Uyt - gt2/2
= 25(3)1/2*4/2 - g*16/2
= 50(3)1/2 - 80
=6.6 m
u can get this result directly by using equation of trajectory ,
Y(height) = Xtan@[ 1- X/R ]
R is horizontal range , X is horizontal distance from point of projection & @ is angle of projection
y=xtanα-g*x2/2u2cos2 α
hence on solving you will get a as answer
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