Question

A soccer ball kicked by a Real Madrid midfielder leaves the player's foot at a speed of 48 mph at an angle θ above the horizontal. Find the angle θ, in radians, that will maximize the distance traveled downrange. (Use g = 32 ft/sec2.)

Answer 1

For maximum range, θ = π/4  

Step-by-step explanation:

Consider the vertical motion of ball,  

    We have equation of motion v = u + at  

    Initial velocity, u = u sin θ  

    Final velocity, v = -u sin θ  

    Acceleration = -g  

Substituting  

    v = u + at  

   -u sin θ = u sin θ - g t  

   $t=\frac{2usin\theta }{g}$  

   This is the time of flight.  

Consider the vertical motion of ball till maximum height,  

We have equation of motion v² = u² + 2as  

   Initial velocity, u = u sin θ

  Acceleration, a = -g  

  Final velocity, v = 0 m/s  

Substituting  

v² = u² + 2as  

0² = u²sin² θ + 2 x -g x H

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

This is the maximum height reached,  

Consider the horizontal motion of ball,  

Initial velocity, u = u cos θ  

Acceleration, a =0 m/s²  

Time, $t=\frac{usin\theta }{g}$  

Substituting  

s = ut + 0.5 at²  

  $s=ucos\theta \times \frac{2usin\theta }{g}+0.5\times 0\times (\frac{2usin\theta }{g})^2\\\\s=\frac{2u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{g}$  

This is the range.  

For maximum range

            sin 2θ = 1

                  2θ = π/2

                  θ = π/4  

For maximum range, θ = π/4  

Answer 2

You can use the fact that the range(distance traveled downrange) of a projectile(height leveled to ground to which it falls) with angle θ is directly proportional to sin of 2θ

The angle θ, in radians, that will maximize the distance traveled downrange is  $\dfrac{\pi}{4}^c$

What is the range of a projectile which was projected from the same level where it falls, at an angle θ?

Suppose the gravitational acceleration be g, and the initial velocity of the projected object be v, then we have;

$d = \dfrac{v}{g} \times \sin(2\theta)$

The range, or distance traveled downrange is the horizontal distance traveled by that projected object.

How to find the angle at which a projected object travels maximum distance?

If you see the formula for "d", you can see that theta is used as input to sin with being doubled firstly.

sin ratio is maximum if it gets input as $n\pi + \dfrac{\pi}{2}$ for n being an integer.

Taking n = 0, sin ratio is maximum at  $\dfrac{\pi}{2}$ radians, this we need theta to be  $\dfrac{\pi}{4}^c$ (that small c in subscript shows radian), it is because when we will multiply 2 to that theta, we will get $2 \times \dfrac{\pi}{4} = \dfrac{\pi}{2}$

It doesn't matter what the initial speed was, neither the value of gravitational acceleration matters. If you shoot the projectile at 45 degrees or   $\dfrac{\pi}{4}^c$, then the projectile, with the use of v and g, will go to maximum of what it can with that v and g.

Thus,

The angle θ, in radians, that will maximize the distance traveled downrange is  $\dfrac{\pi}{4}^c$

Learn more about projectile range here:

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