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 mammoth sixer. The ball starts at 35 m/s at an angle of 53° 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:
Angle of projection of the ball, α = 53°
Width and height of the bench = 1 m
Initial speed of the ball = 35 m/s
Distance of the first bench from the batsman = 110 m
The batsman strikes the ball 1 m above the ground.
Let the ball land on the n^th bench.
∴ y = (n − 1)    ...(i)
And,

\[x = 110 + n - 1 = 110 + y\]

\[\text{ Now } , \]

\[y = x\tan\alpha - \left( \frac{g x^2 \sec^2 \alpha}{2 u^2} \right)\]

\[ \Rightarrow y = \left( 110 + y \right)\left( \frac{4}{3} \right) - \frac{10 \times \left( 110 + y \right)^2 \left( \sec^2 53^\circ \right)}{2 \times \left( 35 \right)^2} \]

\[ = \frac{440}{3} + \frac{4}{3}y - \frac{250 \left( 110 + y \right)^2}{18 \times \left( 35 \right)^2}\]

\[ = \frac{440}{3} + \frac{4}{3}y - \frac{250 \left( 110 + y \right)^2}{18 \times {35}^2}\]

Solving the above equation, we get:
y = 5
⇒ n − 1 = 5
⇒ n = 6
The ball will hit the sixth bench of the gallery.