Question

If shoaib akhtar is bowling at a speed of 164 km/h and the boundary is at 63 metres distance, then by what force should sachin tendulkar hit the ball to score 6 runs.

Answer 1

basically, Mass of cricket ball is approximately 150 grams.

so, m = 150 g = 0.15 kg

ball hits the Sachin Tendulkar's bat with speed , u = 163 km/h = 163 × 5/18 ≈ 45.27 m/s

motion of cricket after hitting by Sachin is projectile. so, path of ball follow equation, y = xtanθ - gx²/2u²cos²θ

for minimum force, θ ≈ 45°

here, x = 63 m , u = 45.27 m/s

then, y = 63tan45° - 10(63)²/2(45.27)²cos²(45°)

= 63 - (10 × 63 × 63)/( 45.27 × 45.27)

≈ 43.6 m

so, path of projectile , s = √(x² + y²)

= √{(43.6)² + (63)²}

= 76.61 m

now from conservation of energy,

Kinetic energy of ball = workdone by ball

or, 1/2 × 0.15 × (45.27)² = F × 76.61

or, F ≈ 2 N

hence, minimum force should be applied by Sachin Tendulkar = 2N

Answer 2

$\huge\bold\purple{Answer:-}$

basically, Mass of cricket ball is approximately 150 grams.

so, m = 150 g = 0.15 kg

ball hits the Sachin Tendulkar's bat with speed , u = 163 km/h = 163 × 5/18 ≈ 45.27 m/s

motion of cricket after hitting by Sachin is projectile. so, path of ball follow equation, y = xtanθ - gx²/2u²cos²θ

for minimum force, θ ≈ 45°

here, x = 63 m , u = 45.27 m/s

then, y = 63tan45° - 10(63)²/2(45.27)²cos²(45°)

= 63 - (10 × 63 × 63)/( 45.27 × 45.27)

≈ 43.6 m

so, path of projectile , s = √(x² + y²)

= √{(43.6)² + (63)²}

= 76.61 m

now from conservation of energy,

Kinetic energy of ball = workdone by ball

or, 1/2 × 0.15 × (45.27)² = F × 76.61

or, F ≈ 2 N