Question 10 of 18intermediate🔧 ApplyNumerical2 marks

A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?

Correct Answer

The cricketer can throw the ball 50 m high above the ground.

Exercise: EXERCISES | Q: 3.13 | (Page 21)
For More Understanding

Explanation

This question connects two concepts from projectile motion: maximum horizontal range and maximum vertical height. For maximum horizontal range, the projection angle must be 45°, making sin(2θ) = 1. The range formula R = v₀²sin(2θ)/g simplifies to R_max = v₀²/g when θ = 45°. From this, we can find v₀² = R_max × g. For maximum height, the ball must be thrown vertically upward (θ = 90°), where the height formula h_max = (v₀sinθ)²/(2g) becomes h_max = v₀²/(2g). Substituting the value of v₀² from the range calculation gives h_max = R_max/2.

Solution Steps

  1. Step 1: Find the maximum horizontal range formula. For maximum horizontal range, the angle of projection θ = 45°. From the context, Range R = v₀²sin(2θ)/g For θ = 45°, sin(2×452 \times 45°) = sin(90°) = 1 Therefore, R_max = v₀²/g

  2. Step 2: Calculate v₀² from the given range. Given: R_max = 100 m So, v₀²/g = 100 Therefore, v₀² = 100g

  3. Step 3: Find the maximum height formula. For maximum height, the ball must be thrown vertically upward (θ = 90°). From the context, Maximum height h_m = (v₀sinθ)²/(2g) For θ = 90°, sin(90°) = 1 Therefore, h_max = v₀²/(2g)

  4. Step 4: Calculate the maximum height. Substituting v₀² = 100g: h_max = 100g/(2g) = 100/2 = 50 m

PreviousNext