From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 9.40 m/s and angle of 23.0° below the horizontal. It strikes the ground 5.00 s later.

By how much does the ball clear or fall short (vertically) of clearing the crossbar? (Enter a negative answer if it falls short.)

To find the vertical distance by which the ball clears or falls short of clearing the crossbar, we need to determine the vertical displacement of the ball.

We can break down the motion of the ball into its horizontal and vertical components:

Horizontal motion: The initial velocity of the ball has two components: one in the horizontal direction (vx) and one in the vertical direction (vy). The horizontal component (vx) remains constant throughout the motion, while the vertical component (vy) is affected by gravity.

Vertical motion: The ball is subject to the vertical component of the initial velocity (vy) and the acceleration due to gravity (g = 9.8 m/s^2). The equation for vertical displacement is given by:

y = y0 + vy0 * t - 1/2 * g * t^2

Where:
y = vertical displacement
y0 = initial height of the ball above the ground (y0 is given)
vy0 = vertical component of the initial velocity (vy0 = v0 * sin(θ) is given, where v0 = 9.40 m/s and θ = 23.0°)
g = acceleration due to gravity (9.8 m/s^2)
t = time (5.00 s is given)

Let's substitute the given values into the equation and calculate the vertical displacement:

y = y0 + vy0 * t - 1/2 * g * t^2

y = y0 + (v0 * sin(θ)) * t - 1/2 * g * t^2

y = y0 + (9.40 m/s * sin(23.0°)) * 5.00 s - 1/2 * 9.8 m/s^2 * (5.00 s)^2

Now, calculate the vertical displacement (y) and determine if the ball clears or falls short of the crossbar.