Question 4 of 18intermediate🔧 ApplyNumerical3 marks

The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s^-1 can go without hitting the ceiling of the hall?

Correct Answer

The maximum horizontal distance is approximately 150.5 m.

Exercise: EXERCISES | Q: 3.12 | (Chapter: 21)
For More Understanding

Explanation

This problem involves projectile motion with a constraint on maximum height. The ball must not hit the ceiling, so the maximum height of the projectile must be equal to or less than 25 m. We use the formula for maximum height to find the required angle of projection, then calculate the corresponding horizontal range.

Solution Steps

  1. Step 1: Given values: Speed v₀ = 40 m s⁻¹, Maximum height allowed hₘ = 25 m, g = 9.8 m s⁻²

  2. Step 2: Find the angle of projection using the maximum height formula: hₘ = (v₀ sin θ)²/2g

  3. Step 3: Substituting values: 25 = (40 sin θ)²/(2×9.82 \times 9.8)

  4. Step 4: Solving: 25×19.625 \times 19.6 = 1600 sin²θ → sin²θ = 490/1600 = 49/160

  5. Step 5: Therefore, sin θ = 7/160\sqrt{160} = 7/(410\sqrt{10}) ≈\approx 0.5534

  6. Step 6: Calculate cos θ: cos θ = 1−sin2θ\sqrt{1 - sin²θ} = 111/160\sqrt{111/160}

  7. Step 7: Find sin 2θ: sin 2θ = 2 sin θ cos θ = 2 × (7/160\sqrt{160}) × (111\sqrt{111}/160\sqrt{160}) = 14111\sqrt{111}/160 ≈\approx 0.922

  8. Step 8: Calculate horizontal range: R = (v₀² sin 2θ)/g = (1600×0.9221600 \times 0.922)/9.8

  9. Step 9: Final answer: R = 1475.2/9.8 ≈\approx 150.5 m

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