Question

In the game with Green-Bay the Bears lost the game with a touchdown of 40 m range. The hangtime of the ball was 3.2 sec.
a. what was the initial velocity of the ball?
b. what was the maximum height of the ball?
c. what was the final velocity of the ball ?
(physics)​

Answer 1

The projectile launch allows to find the results for questions about the movement are:

   a) The initial velocity is: vo = 20 m/s with an angle of θ = 51.4º

   b) The maximum height of the ball is:  y = 12.5 m

   c) The final velocity is: vo = 20 m/s with θ = 308.3º

The launch of projectiles is an application of kinematics where there is no acceleration on the x axis and the y axis is the gravity acceleration.

In the attachment we see a diagram of movement, the x-axis is horizontal and the y-axis is vertical.

They indicate that the range was x = 40 m and the ball was in the air t=3.2s.

a) Let's find the horizontal velocity, as there is no acceleration.

         x = v₀ₓ t

         v₀ₓ = $\frac{x}{t}$x / t

let's calculate

         v₀ₓ = $\frac{40}{3.2}$  

         v₀ₓ = 12.5 m / s

We look for the vertical speed, when the ball reaches the destination its height is zero.

          y = $v_o_y$ t - ½ g t²

          0 = $v_o_y$ t - ½ g t²

          0 = t ( $v_o_y$ - ½ g t)

The solution to this equation is:

           t = 0 s         Exit time.

           $v_o_y$ - ½ g t = 0

           $v_o_y$  = ½ g t

It indicates the time in the air that, because it is a scalar, is the same for the two movements.

Let's calculate

           $v_o_y$  = ½ 9.8 3.2

            $v_o_y$  = 15.68 m / s

We use the Pythagoras' theorem for the initial velocity modulus.

          v₀ = $\sqrt{v_{ox}^2 + v_{oy}^2 }$  

          v₀ = $\sqrt{12.5^2 + 15.68^2}$  

          v₀ = 20 m / s

We use trigonometry for the angle.

        tan θ = $\frac{v_o_y}{v_o_x}$  

        θ = tam⁻¹ $\frac{v_o_y}{v_o_x}$  

        θ = tan⁻¹  $\frac{15.68}{12.5}$  

        θ = 51.4º

b) For the maximum height the vertical speed is zero.

           $v_y^2 = v_{oy}^2 -2g y$  

           0 = $0 = v_{oy}^2 -2 g y$  

           y = $\frac{v_{oy}^2}{2g}$

Let's calculate

           y = $\frac{15.68^2}{2 \ 9.8}$

           y = 12.5 m

C) The final velocity of the ball.

Since the x-axis there is no acceleration, the velocity is constant.

          vₓ = 12.5 m / s

Let's find the final vertical speed.

         $v_y = v_{oy} - g t$

         $v_y$ = 15.68 - 9.8 3.2

         $v_y$ = - 15.68 m / s

We are looking for the module.

         v = vₓ² + $v_y^2$  

speed gives the same value.

         v = 20 m / s

We use trigonometry for the angle.

        θ'= tan⁻¹  $\frac{-15.68}{12.5}$

        θ' = -51.66º

This angle measured from the positive side of the x-axis counterclockwise is:

       θ = 360 - θ'

       θ = 360 - 51.66

       θ = 308.3º

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

   a) The initial velocity is: vo = 20 m / s with an angle of θ = 51.4º

   b) The maximum height of the ball is: y = 12.5 m

   c) The final velocity is: vo = 20 / s with θ= 308.3º

Learn more here: brainly.com/question/15063198