During a baseball game, a batter hits a popup to a fielder 64 m away.

The acceleration of gravity is 9.8 m/s
2
.
If the ball remains in the air for 6.7 s, how
high does it rise?
Answer in units of m

Dx = Xo*6.7 = 64.

Xo = 9.55 m/s = Hor. component of initial velocity.

Tr = Tf = 6.7/2 = 3.35 s. = Rise and Fall time.

Y = Yo + gt = 0.
Yo - 9.8*3.35 = 0
Yo = 32.83 m/s. = Ver. component of
initial velocity.

h = Yo*t + 0.5g*t^2.
h = 32.83*3.35 -4.9*(3.35)^2 = 55 m.

To find the height that the ball rises, we can use the formulas of projectile motion, considering that the vertical motion is affected by gravity.

Let's break down the problem and find the necessary information:

Given:
- Distance covered horizontally (d): 64 m
- Time of flight (t): 6.7 s
- Acceleration due to gravity (g): 9.8 m/s^2

We need to find the height the ball rises (h).

First, we can find the initial vertical velocity (u_y) using the formula:

u_y = h / t

Now, we know that the ball's velocity in the vertical direction decreases due to the acceleration of gravity. The final vertical velocity (v_y) can be calculated using the formula:

v_y = u_y - g * t

Since the ball reaches its maximum height when its final vertical velocity becomes zero, we can write:

0 = v_y - g * t

Solving for u_y:

u_y = g * t

Now, we can use the equation of motion to find the height (h):

h = (u_y * t) - (0.5 * g * t^2)

Substituting the known values:

h = (g * t^2) - (0.5 * g * t^2)

Simplifying:

h = 0.5 * g * t^2

Plugging in the values:

h = 0.5 * 9.8 * (6.7)^2

Calculating:

h ≈ 0.5 * 9.8 * 44.89
h ≈ 219.19

Therefore, the ball rises approximately 219.19 meters.