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5. A quarterback throws the football to a stationary receiver who is 31.5 m
down the field. If the football is thrown at an initial angle of 40.0° to the
ground, at what initial speed must the quarterback throw the ball for it
to reach the receiver? What is the ball's highest point during its flight?

Respuesta :

The projectile launch equations allow to find the results for the questions about the movement of the ball are:

  • The initial velocity is:   vâ‚€ = 17.7 m / s.
  • The maximum height is:   y = 16 m.

Given parameters

  • Horizontal distance x = 31.5 m
  • Launch angle tea = 40º

To find

  • The initial speed.
  • Maximum height.

Projectile launching is an application of kinematics, where on the x-axis there is no acceleration and on the y-axis is the gravity acceleration.

The range is the distance traveled for the same departure height, see attached.

.

          R =[tex]\frac{v_o^2 \ sin 2\theta}{g}[/tex]  

         [tex]v_o^2 = \frac{ g R}{sin 2 \theta }[/tex]  

Let's calculate.

          v₀² = [tex]\frac{9.8 \ 31.5}{sin \ (2 \ 40)}[/tex]9.8 31.5 / sin (2 40.0)

          [tex]v_o = \sqrt{313.46}[/tex]o = ra 313.46

          v₀ = 17.7 m / s

At the point of maximum height the vertical speed is zero.

          v² = v₀² - 2 g y

          0 = v₀² - 2g y

          y = [tex]\frac{v_o^2}{2g}[/tex]  

Let's calculate.

         y = [tex]\frac{17.7^2}{2 \ 9.8}[/tex]  

         y = 16 m

In conclusion, using the projectile launch equations we can find the results for the questions about the movement of the ball are:

  • The initial velocity is: vâ‚€ = 17.7 m / s
  • The maximum height is:  y = 16 m.

Learn more here:  brainly.com/question/10903823

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