Afootball 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?

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  

     

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

 

This is the maximum height reached,  

Consider the horizontal motion of ball,  

  Initial velocity, u = u cos θ  

  Acceleration, a =0 m/s²  

  Time,  

Substituting  

  s = ut + 0.5 at²  

   

This is the range.  

Here θ = 45° and S = 60.6 m

Substituting

           

Initial speed of the ball is 24.38 m/s

Hello,

Let's assume CD=h, BD=x.

tan 24°=h/x ==>h=x*tan 24°

tan 16°=h/(7600+x)==>h=tan 16° *(7600+x)

==>x tan 24°=tan 16°*(7600+x)
==>x=(7600*tan 16°)/(tan 24°-tan 16°)

h=x*tan 24= (7600*tan 16°*tan 24°)/(tan 24°-tan16°)

b) CD=h=6122,2303...≈6122 (feet)

a) BC=h/sin 24°=15052,0746448...≈15052 (feet)

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