So, that's why they love cricket

Classical Mechanics Level 4

In a cricket match, a 2-meter-tall pace bowler throws a yorker ball with a horizontal velocity of 20 m/s 20\text{ m/s} . The batsman hits the ball with twice the horizontal velocity of the ball at an angle of 45 ∘ {45}^\circ with the horizontal. The ball lands just in front of a fielder, who kicks the ball with his foot at an angle of 30 ∘ {30}^\circ with the horizontal in such a way that it lands just in front of the foot of the bowler. Find the velocity with which the fielder kicked the ball.

Details and Assumptions:

  • The air friction is negligible.
  • The positions of the batsman, bowler, and fielder are collinear and the batsman hits the ball in the collinear line.
  • The bowler throws the ball by stretching his hand just above the top of his head.
  • The bowler does not move from his position after releasing the ball from the crease line.
  • Take g = 10 m/s 2 g=10{\text{ m/s}}^2 the acceleration due to gravity.
  • Round your answer ( ( in m/s ) \text{m/s}) off to the nearest integer.


The answer is 41.

Ashish Menon by Ashish Menon , 20, India

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1 solution

Ashish Menon
Jul 10, 2016

The range of the projectile of the ball from the pacemans hand is u 2 h 2 g = 20 × 2 × 2 10 = 40 10 u\sqrt{\dfrac{2h}{2g}}\\ \\ = 20×\sqrt{\dfrac{2×2}{10}}\\ \\ = \dfrac{40}{\sqrt{10}}

The range of projectile of the ball hit by batsmen = u 2 sin ⁡ 2 θ g = 40 × 40 sin ⁡ 90 ° 10 = 160 \dfrac{u^2\sin2\theta}{g}\\ \\ = \dfrac{40×40\sin {90}^°}{10}\\ \\ = 160 .

Range of the projectile of the ball kicked by fielder = 160 − 40 10 160 - \dfrac{40}{\sqrt{10}} .
Now, range = u 2 sin ⁡ 2 θ g \dfrac{u^2\sin2\theta}{g} .

160 − 40 10 = u 2 sin ⁡ 60 10 160 10 − 40 10 = u 2 3 20 u 2 = 1701.449 u = 1701.449 = 41.2489 ≈ 41 \begin{aligned} 160 - \dfrac{40}{\sqrt{10}} & = \dfrac{u^2\sin60}{10}\\ \dfrac{160\sqrt{10} - 40}{\sqrt{10}} & = \dfrac{u^2\sqrt{3}}{20}\\ u^2 & = 1701.449\\ u & = \sqrt{1701.449}\\ & = 41.2489 & \approx \color{#3D99F6}{\boxed{41}} \end{aligned}

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