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 out how much the ball clears or falls short of clearing the crossbar, we need to determine the vertical position of the ball when it reaches the ground.
First, let's break down the initial velocity into its vertical and horizontal components. The vertical component can be found using the equation v_y0 = v_0 * sin(θ), where v_y0 is the initial vertical velocity, v_0 is the initial velocity, and θ is the angle below the horizontal. Substituting the given values: v_y0 = 9.40 m/s * sin(23.0°).
Next, we can calculate the time it takes for the ball to reach the ground using the equation y = y_0 + v_y0 * t + (1/2) * g * t^2, where y is the final vertical position, y_0 is the initial vertical position, v_y0 is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (approximately -9.8 m/s^2). Substituting the given values, y = 0 (as the ground is the reference point), y_0 = y0, v_y0, and t = 5.00 s, we can solve for y0.
Once we have y0, we can determine the difference between y0 and the height of the crossbar to find out how much the ball clears or falls short.
Let's perform these calculations to find the answer.