Question

The maximum height attained by a projectile when thrown at an angle $\theta $ with the horizontal is found to be half the horizontal range. Then $\theta $ =

• A. $tan^{-1} \frac {1}{2} $
• B. $\frac {\pi} {4} $
• C. $\frac {\pi} {6} $
• D. $tan^{-1} (2)$

Answer 1

The Correct Option is D

Maximum height, $H_{0}=\frac{u^{2} \sin _{2} \theta}{2 g}$
Range, $ R=\frac{u^{2} \sin 2 \theta}{g}$
Given, $ H_{0}=\frac{R}{2}$
$\therefore \frac{u^{2} \sin ^{2} \theta}{2 g}=\frac{u^{2} 2 \sin \theta \cos \theta}{2 g}$
$\Rightarrow \sin \theta=2 \cos \theta$
$\Rightarrow \tan \theta=2$
$\therefore \theta=\tan ^{-1}(2)$

Answer 2

The angle at which the projectile is thrown is θ

Let its initial velocity be u. 

As the projectile is dependable on gravity, the acceleration is downwards.

Let's consider the maximum height acquired by the projectile is H and the horizontal range of the projectile is R.

It is given that 

H=\(\frac{R}{2}\)

 

First, we will find the expressions for H and R in the respect of u and θ and then equate them according to equation (i).

Resolve the velocity vector u horizontally (ux) and vertical (uy) components.

U_x = ucosθ

U_y = usinθ

When the projectile is placed at its maximum height, its displacement is equivalent to H and also its velocity at this point is equal to 0.

Here, we can use the kinematic equation, 

2as=v²−u² for the of u in the vertical direction and the values of a, s, u, and v as g, H, u_y=usinθ and 0 respectively.

Therefore we get,

2(−g)(H)=0²−(usinθ)²

⇒−2gH=−u²sin²θ

Therefore, 

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

For R we can apply the horizontal component of u i.e. u_x=ucosθ

Here, since there is no force to affect the horizontal velocity, its acceleration in this direction will be 0.

 

So, we can simply use the formula, 

displacement= velocity × time

Let that time be T.(T is the time of the projectile to hit the ground)

When the ball hits the ground, the vertical displacement of the projectile is 0.

To find T, we may use the kinematic equation, s=ut+\(\frac{1}{2at^2}\)

Here, s=0, u=u_y=usinθ, t=T, and a = -g.

Substitute the respective values in equation (iii).

Therefore, we get,

0=usinθT+\(\frac{1}{2}\)(−g)T²

⇒\(\frac{1}{2}gT^2\)=usinθT

Divide both sides by T.

⇒\(\frac{1}{2gT}\)=usinθ

⇒T=\(\frac{2usin\theta}{g}\)

And we know that 

R=uxT=ucosθT

Therefore, 

R=\(\frac{2u2sin\theta cos\theta}{g}\) ……… (iv)

From (i), (ii), and (iii) we get,

\(\frac{u^2sin2\theta}{2g} = \frac{2u2sin\theta cos\theta}{\frac{g}{2}}\)

⇒ \(sin2\theta = 2sin\theta cos\theta\)

Subtracting 

2sinθcosθ on both the side, we get

⇒sin2θ−2sinθcosθ=0

taking sinθ common, we will get

⇒sinθ(sinθ−2cosθ)=0

⇒sinθ=0  which is not possible or

⇒sinθ/cosθ=2

⇒tanθ=2 ⇒ θ=tan⁻¹(2)

Hence, the correct option is D.

 

Answer 3

Maximum height, H = \(\frac{u^2 sin2\theta}{2g}\)

Range, \(R = \frac{u^2sin2\theta}{g}\)

Given, \(H=\frac{R}{2}\)

∴ \(\frac{u^2sin2\theta}{2g} =\frac{u^2 2sin\theta cos\theta}{2g}\)

or, \(sin\theta = 2cos\theta\)

or tanθ=2

or θ=tan⁻¹(2)

 

The correct option will be (D)