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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```
8 years ago

## Answers : (4)

```							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

```
8 years ago
```							y=xtanα-g*x2/2u2cos2 α
hence on solving you will get a as answer
```
8 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 knowIt can directly come by equation of trajectory
```
one year 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
```
10 months ago
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