Question

3) What should be the angle of projection for the horizontal range of a projectile to be maximum?
(a) 90 0 (b) 1800 (c) 450 (d) 00

Answer 1

Let Velocity is V and the angle of projection with horizontal is θ.

∴Range, R=g2V2sinθcosθ=gV2sin2θ

⇒dθdR=gV22cos2θ=0 ⇒2θ=90o ⇒θ=45o

⇒dθ2d2R=−gV24sin2θ, at θ=45o $$\Rightarrow \dfrac{d^2R}{d\theta^2}=-\dfrac{4V^2}{g}<0$$

Hence, at θ=45o , Range is maximum.

Answer 2

Correct Question:

What should be the angle of projection for the horizontal range of a projectile to be maximum?

(a) 90° (b) 180° (c) 45° (d) 0°

Answer:

The angle of projection (θ) for which the angle of projection to have a maximum range will be 45 °

Given:

1. Range formula is 2 u² sinθ cos θ / g

Explanation:

$\rule{300}{1.5}$

From the formula we know,

⇒ R = 2 u² sinθ cosθ / g

Here,

• R is range
• u is Initial velocity.
• θ is Angle of projection.
• g denotes acceleration due to gravity.

Now,

⇒ R = 2 u² sinθ cosθ / g

⇒ R =  u² × ( 2 sinθ cosθ ) / g

⇒ R = u² × sin 2θ / g  ∵ [ 2 sinθ cosθ = sin 2θ ]

⇒ R = u² sin 2θ / g

Here, u and g are constant.

Therefore,

For having maximum the range the angle should be more.

Now,

We know that maximum value of sin angle is 1.

Therefore,

⇒ sin 2θ = 1

⇒ sin 2θ = sin 90 ∵ [ sin 90 = 1 ]

⇒ 2 θ = 90 °

⇒ θ = 90 ° / 2

⇒ θ = 45 °

⇒ θ = 45 °

∴ The angle of projection (θ) for which the angle of projection to have a maximum range will be 45 °.

$\rule{300}{1.5}$