Project 2

Classical Mechanics Level 1

There are 5 boys who throw the ball at different angles from the ground.

A throws at $30^\circ$ .
B throws at $45^\circ$ .
C throws at $15^\circ$ .
D throws at $90^\circ$ .
E throws at $60^\circ$ .

Whose ball will cover maximum horizontal distance?

Image Credit: Flickr Butch Siemens .

A B C D E

6 solutions

Parag Zode
Jun 28, 2015

Range of projectile is:- $R = \dfrac{v^2 sin2 \theta}{g}$ We can see that the range will be maximum when the value of $sin2 \theta$ is highest i.e. $2{\theta} = 90^{\circ}$ . So, $\theta$ must be equal to $45^{\circ}$

Thank you, can you tell me which field of maths/physics should i refer to to learn this kind of stuff?

Bilal Akmal - 5 years, 10 months ago

You will get this kind of stuff to read in chapter Kinematics.. and by doing a bit of revision of vector algebra.. Hope this helps.. $\ddot\smile$

Parag Zode - 5 years, 10 months ago

Thanks. It does :)

Bilal Akmal - 5 years, 10 months ago

With air resistance, doesn't a 30 degree angle go further than a 45 degree angle?

Hugh Goatcher - 5 years, 6 months ago

No it doesn't .. Neglect air resistance

Parag Zode - 5 years, 6 months ago

Adrian Peasey
Jun 30, 2015

Consider the general case for projectile motion from the origin.

In the horizontal direction we have $s=x$ , $u=v\cos\theta$ , and $a=0$ .

In the vertical direction we have $s=0$ , $u=v\sin\theta$ , and $a=-g$ (neglecting the height of the boy).

Using $s=ut+\frac{1}{2}at^2$ we obtain:

$x=vt\cos\theta+0\Rightarrow t=\frac{x}{v\cos\theta}$

and

$0=vt\sin\theta-\frac{1}{2}gt^2 \Rightarrow t=\frac{2v\sin\theta}{g}|t\neq0$

Equating these we have: $\frac{2v\sin\theta}{g}=\frac{x}{v\cos\theta}\Rightarrow x=\frac{v^2 \sin(2\theta)}{g}$

Assuming all 5 boys throw with the same speed this is maximised for: $\sin(2\theta_{max})=1\Rightarrow 2\theta_{max}=\sin^{-1}(1)=90\Rightarrow \theta_{max}=45$

So B throws furthest.

@Adrian Peasey Thanks. Great derivation explaining the basics of the projectile motion. One questions - in case of horizontal component, why is a(acceleration)=0? I guess it because gravity is a vertical force, with its horizontal component being zero. But intuitively, one would think gravity acts even on horizontal motions.

Ajit Deshpande - 5 years, 10 months ago

When we consider gravity we are actually using the component in the horizontal and vertical directions. The $a=0$ and $a=-g$ statements are actually $a=-g\sin\theta$ and $a=-g\cos\theta$ respectively, where $\theta=0$ is the angle from the vertical.

Adrian Peasey - 5 years, 10 months ago

Eric Osborn
Jul 28, 2015

Seems like there should be some mention that they're throwing with equal force. Not that one could really get anywhere assuming otherwise, just seems like an oversight not to mention that