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