1. At what angle of release did the sepak takraw ball travel farthest? 2. How will you compare the range (horizontal distance) travelled by the ball at angles of 25º and 65º? How about at angles of 40º and 50º? 3. At what angle of release did the ball reach the maximum height? 4. How will you compare the height reached by the ball at 25º, 45º and 65º? 5. How will you explain the relationship between the angle of release to the height of the projectile? How about angle of release and range before it reaches the farthest distance?

1) θ = 45,  b)  R₄₀ = R₅₀ = (v₀² / g) 0.985,  R₂₅ = R₆₅ = (v₀² / g) 0.766

3)  θ = 90º, d)   y₂₅ = (v₀ / 2g)   0.179, y₄₅ = (v₀ / 2g) 0.5, y₆₅ = (v₀ / 2g) 0.821

Explanation:

This is a projectile throwing exercise, let's use the relations

         R =

1) the angle for a maximum reach (R) is θ = 45

         R = v₀²/g

2) let's compare the scopes

θ = 25º

          R₂₅ = (v₀² /g) sin 2 25

          R₂₅ = (v₀² /g)  0.766

θ = 65º

         R₆₅ = (v₀² / g)  sin 2 65

         R₆₅ = (v₀² / g) 0.766

θ = 40º

        R₄₀ = (v₀²/g)  sin 2 40

        R₄₀ = (v₀²/ g) 0.985

θ = 50

        R₅₀ = (v₀²/g)  sin 2 50  

        R₅₀ = (v₀² / g) 0.985

we can see that it has the same scope

         R₄₀ = R₅₀ = (v₀² / g) 0.985

         R₂₅ = R₆₅ = (v₀² / g) 0.766

3) the maximum height is reached when the vertical speed is zero

          v_y = 0

         y =

         y = (v₀² /2g) ( sin θ)²

therefore the height is maximum for θ = 90º

4) θ = 25º

         y₂₅ = (v₀ / 2g)   (sin 25)²

         y₂₅ = (v₀ / 2g)   0.179

θ = 45º

         y₄₅ = (v₀ / 2g) (sin 45)²

         y₄₅ = (v₀ / 2g) 0.5

θ = 65º

         y₆₀ = (v₀ / 2g) (sin 65)²

         y₆₅ = (v₀ / 2g) 0.821

5) It is appreciated that as the launch angle increases the height increases, the maximum is when the movement is vertical

For the launch angle, the reach has a maximum for the angle of 45º, this is due to the symmetry of the sine function around this value.

correct it is c.) b b