Abaseball pitcher throws a fastball at 98 mph. he also has a slower "changeup" pitch, 75 mph, which he throws identically, same windup and same release angle from shoulder height. which pitch gets to home plate with a flatter i. e, velocity vector closer to horizontal

For these reasons at 98 mph the path is straighter

Explanation:

To solve this problem we are going to use the kinematic equations, specifically those of projectile launches, let's calculate the distances that the ball travels

          X = Vox t

          Y = Yo +  Voy t - ½ g t²

They tell us that the only parameter that changes is the speed, so the distance to the plate is known

          t = Vox / x

We replace

         Y- Yo = Voy (Vox / X) - ½ g (Vox / x)²2

       Y -Yo = Vo² sinθ cos θ / x - ½ g Vo² sin²θ / x²

        Y -Yo = Vo² (sinθ cosθ / x - ½ g sin²θ / x²)

The trajectory will be flatter when Y is as close as possible to Yo, when examining the right side of the equation, the amount in Parentheses is constant and to what they tell us that the angles and the distance the plate does not change.

Consequently, of the above, the only amount changes is the initial speed if it increases the square of the same increases, so that the height Y approaches the height of the shoulder, that is, DY decreases. For these reasons at 98 mph the path is straighter

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.