Question

When a stone is thrown at an angle of 45°, with the horizontal it rises to a maximum

height of 10m. What is its horizontal range?

Answer 1

Answer: 40 m

Explanation:

The horizontal range is given by:

$R=\frac{u^2sin2\theta}{g}$

The maximum height achieved by the projectile is given by:

$H = \frac{u^2 sin^2\theta}{2g}$

Where, θ is angle of projectile, u is the initial speed with which projectile is thrown and g is the acceleration due to gravity.

Let us the ratio of two:

$\frac{R}{H} = \frac{\frac{u^2sin2\theta}{g}}{\frac{u^2 sin^2\theta}{2g}} =\frac{2sin2\theta}{sin^2\theta}=\frac{2\times 2sin\theta cos\theta}{sin^2\theta}=4cot\theta$

We are given Θ=45°

H = 10 m

⇒R = 4 ×cot 45°×10 m = 40 m

Hence, the horizontal range of the stone thrown at an angle of 45° is 40 m

Answer 2

Answer: 40 m

Explanation:

The horizontal range is given by:

The maximum height achieved by the projectile is given by:

Where, θ is angle of projectile, u is the initial speed with which projectile is thrown and g is the acceleration due to gravity.

Let us the ratio of two:

We are given Θ=45°

H = 10 m

⇒R = 4 ×cot 45°×10 m = 40 m

Hence, the horizontal range of the stone thrown at an angle of 45° is 40 m