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a batsman hits a ball at an angle of 53deg with horizontal at a height of 1m above the ground towards the audience with a velocity of 35m/s. If the distance of first bench from him is 110m and each bench has same height and width then find which bench will be struck by the ball
Hello Neeraj
Now the ball is struck by the bat at a height of 10 m...and the angle is 53deg ...so it becomes a case like ...a projectile is thrown from a tower ....
Now Velocity is 35 m/s ...it is at an angle of 53 with the horizontal..so it will have two components ..one in horizontal and one in vertical..
Horizontal component = 35.cos(53) = 21 m/s
Vertical component = 35.sin(53) = 28 m/s
Now vertical velocity = 28 m/s is responsible to bring the ball back to the ground .
and horizontal component will let it to move forwar to the bench ...
So ...Let us calculate what time is taken by the ball to hit the ground...which is due to v=28m/s
h = ut + 1/2 a t ^2
-1 = 21(t) - 5(t)^2
5t^2 -21 t - 1 = 0
t = [21 + root( 461)]/10
t= 4.2 sec .
Now in this much time the ball will hit the ground...horizontal velocity v = 28 m/s is responsible for the horizontlal move of the ball....so let us calculate how far the ball will go and strike...
t = 4.2 sec ...v = 28 m/s
so distance in horizontal direction (range) = 4.2 X 28 = 117.6 m
The First bench is at 110 ..
So ball strike something ...7.6 m behind the first bench
Given spacing and height of the bench is same ..i.e = 1 m
Then ...it will hit to the eighth bench.....because spacing is 1 m..and extran error is 0.6 ...so it is just 60 cm < 1m
Hence after 7 m...0.6 proves that eight bench is the answer
--------------------------------110 ---------------( 1 bench)( ) ( ) ( ) ( ) ( ) ( ) ( )
..................7.6..............................
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