The benches of a gallery in a cricket stadium are 1 m wide and 1 m high. A batsman strikes the ball at a level one meter above the ground and hits a mammoth sixer. The ball starts at 35 m/s at an angle of `53^0` with the horizontal. The benches are perpendicular to the plane of motion and the first bench is 110 m from the batsman. On which bench will the ball hit?

Explanation

Given, `theta = 53^0
Therefore, `cos 53^0 = 3/5
`sec^2 theta = 25/9`
and ` tan theta = 4/3`
Suppose the ball lands on ninth bench
`:. y=(n-1)1……….i.`
[ ball starting point 1 m above ground ]

Again ` y= x tan theta = ((gx^2 sec^2 theta)/(2u^2)) `
`rarr y= (110+y) (4/3) - (10(110+y)^2(25/9))/(2xx(35)^2)`
`x= 100+n-1=100+y[Using i. ]`
`=440/3+4/3 y (250(100+y)^2)/(18xx(35)^2)`
`440/3+4/3 y (250*110+y)^2)/(18xx35^2)`
From the equation, y cane calculatd as
y=5 ltbr. `rarr n-1 = 5`
`rarr n=6` ltbr. The ball will drop in sixth bench.