In a criket match a batsman hits a six . Ball is hit 1 m above the ground level from a point which is 110m away from the benches of stadium . Ball leaves the bat with a speed `35 m//s` and at an angle `53^(@)` above the horizontal. Height and width of benches are 1 m each . Assume that benches are perpendicular to the plane of motion of the ball. Which bench will the ball hit ?

Explanation

Situation described in question is shown in the following figure . Initial point of motion of ball is marked as point O and is selected as the origin of the coordinate system . X-axis is horizontal and Y-axis is vertically upward.

Benches are 1 m high and 1 m wide and if we join the edge of the benches, then we get a plane inclined at an angle `45^(@)` with the horizontal . On this imaginary palne , length of each step is `sqrt2m` . Let the ball hit on this plane at a distnce x from its bottom . Components of displacement of ball can be written as follows :
`S_(x)=110+x cos 45^(@)=110+(x)/(sqrt2),`
`S_(y)x sin 45^(@)=(x)/(sqrt2),`
Components of initial velocity are shown in the figure .
`u_(x)=21m//s,u_(y)=28m//s,`
We can write the components of acceleration as follows:
`a_(x)=0,a_(y)=-9.8m//s^(2)`,
X-axis, `S_(x)=u_(x)t+(1)/(2)u_(x)t^(2) rArr 110+(x)/(sqrt2)=21t`
Y-axis : `S_(y)=u_(y)t+(1)/(2)a_(y)t^(2) rArr(x)/(sqrt2)=28t-4.9t^(2)`
Substituting value of `(x)/(sqrt2)` from equation (ii) in equation
(i), we get the following:
`rArr 110+28t-4.9t^(2)=21t`
`rArr 4.9t^(2)-7t-110=0`
`rArr t=(7pmsqrt(49+4xx4.9xx110))/(9.8)`
`rArr t=(7 pm 7sqrt45)/(9.8)`
We shall consider the positive sigh only and rearrange as follows:
`rArr t=(1+3sqrt5)/(1.4)=5.5s`
Substituting above result in equation (1),
`rArr 110 +(x)/(sqrt2)=21xx5.5 rArr x=5.5 sqrt2`
We have already discussed that each step is `sqrt2` m, hence
`(x)/(sqrt2)=5.5`
We can that it is more than `5^(t"th")` step , hence ball hits the `6^("th")` step.