Question

A football player punts the ball at a 45.0º angle. When the ball returns to the ground, it will have a horizontal displacement of 60.6 m. What is the initial speed of the ball?

Answer 1

Initial speed of the ball is 24.38 m/s

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.  

Here θ = 45° and S = 60.6 m

Substituting

           $s=\frac{u^2sin2\theta}{g}\\\\60.6=\frac{u^2\times sin(2\times 45)}{9.81}\\\\u=24.38m/s$

Initial speed of the ball is 24.38 m/s