# 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

509 Points
13 years ago

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

Black Widow
36 Points
13 years ago

y=xtanα-g*x2/2u2cos2 α

hence on solving you will get a as answer

Samuel Garry
61 Points
5 years ago
It's also is 8.2 metres that is a option as it is the most closest 6.6 metres rest I don't know
It can directly come by equation of trajectory
Anuj Tapare
13 Points
5 years ago
Take g= 9.8 so we can get ans 8.2 we have to take 9.8 due this is a real type que , and we all know that we assume 10 instead of 9.8