The pitcher throws a pitch from the mound toward home plate 18.4m away. The ball is released horizontally from a height of 1.8m with a velocity of 32m/s. Find the height above ground when it gets there.

time of flight: 18.4/32 sec

hf=hi-4.9t^2 In the vertical
hf=1.8-4.9(t you compute it)^2

To find the height above the ground when the ball reaches home plate, we can use the equations of motion in two dimensions: one for horizontal motion and one for vertical motion.

First, let's consider the horizontal motion. Since the ball is released horizontally, there is no acceleration in the horizontal direction. Therefore, the horizontal component of the ball's velocity remains constant throughout its motion. We can find the time it takes for the ball to reach home plate using the horizontal distance and velocity:

Horizontal distance (d) = 18.4 m
Horizontal velocity (v) = 32 m/s

The time (t) it takes for the ball to reach home plate can be found using the equation:
d = v * t

Plugging in the known values:
18.4 = 32 * t

Solving for t:
t = 18.4 / 32 = 0.575 s

Now that we know the time it takes for the ball to reach home plate, we can use the vertical motion equation to find the height above the ground when it gets there.

In vertical motion, there is a constant acceleration due to gravity acting downward. The initial vertical velocity is 0 (since the ball is released horizontally). The vertical displacement (h) can be found using the equation:

h = v0 * t + (1/2) * g * t^2

Where:
v0 = initial vertical velocity = 0 m/s
t = time taken to reach home plate = 0.575 s
g = acceleration due to gravity = 9.8 m/s^2

Plugging in the values:
h = 0 * 0.575 + (1/2) * 9.8 * (0.575)^2

Simplifying the equation:
h = 0 + 0.5 * 9.8 * 0.331

Calculating the result:
h = 1.6 m

Therefore, the height above the ground when the ball reaches home plate is 1.6 meters.