Question

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 metre above the ground and hits a manmoth 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 ?

Answer 1

Given, $\theta ={53}^0$

$\therefore ,cos{53}^0=\frac{5}{3}$

$se{c}^2\theta =\frac{25}{9}$

and $tantheta=\frac{4}{3}$

$\therefore y=(n-1)1$ ....(i)

{ball starting point 1 m above ground}

Again y = $xtan\theta -\left(\frac{g{x}^{2}se{c}^2\theta }{2a^2}\right)$

$\Rightarrow y=(110+y)\left(\frac{4}{3}\right)-\frac{10(110+y{)}^2(25/9)}{2\times (35{)}^2}$

[x= 100 + n -1 =100 +y [Using(i)]
= $\frac{440}{3}+\frac{4}{3}y\frac{250(110+y{)}^2}{18\times \left({35}^2\right)}$
= $\frac{440}{3}+\frac{4}{3}y\frac{250(110+y{)}^2}{18\times {35}^2}$

From the equation, y can be calculated as

y = 5

$\Rightarrow$ n-1=5

$\Rightarrow$ n = 6

The ball will drop in sixth bench