During a baseball game, a batter hits a pop-up to a fielder 83 m away.

The acceleration of gravity is 9.8 m/s^2.
If the ball remains in the air for 6.5 s, how high does it rise?
Answer in units of m

)A batter hits a pop-up to a fielder 83 m away. If the ball remains in the air for 6.5 s, how high does it rise?

Rise time = fall time = 3.25sec.
Vf = 0 = Vo - 9.8(3.25) or Vo = 31.85m/s

d = Vo^2(sin(2µ))/g
----31.85^2(sin(2µ))/9.8 or µ = 26.65º

Then, height h = Vo^2(sin^(x)/2g or
h = 31.85^2(sin(26.65))/2(9.8) = 10.41m

To find the height the ball rises, we can use the kinematic equation:

d = v₀t + 0.5at²

Where:
d is the displacement (change in vertical position)
v₀ is the initial vertical velocity (which would be zero in this case since the ball is hit vertically)
t is the time the ball is in the air
a is the acceleration due to gravity

First, let's find the initial velocity, v₀. Since the ball is hit vertically, the initial velocity in the vertical direction is zero.

Next, we can substitute the given values into the equation.

d = 0 + 0.5 * (-9.8 m/s²) * (6.5 s)²

Simplifying:

d = -0.5 * 9.8 m/s² * (6.5 s)²

d = -0.5 * 9.8 m/s² * 42.25 s²

d = -0.5 * 410.35 m²/s²

d = -205.18 m²/s²

Since displacement (d) cannot be negative in this case, we can take the absolute value to get the magnitude of the displacement:

|d| = 205.18 m²/s²

Now, we have the magnitude of the displacement, but we need to find the height. The height is the vertical distance from the starting point to the peak of the ball's trajectory. Since the ball was hit vertically, the height will be the same as the magnitude of the displacement:

height = |d| = 205.18 m

Therefore, the ball rises to a height of 205.18 meters.