A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 15.0 m/s at an angle 5.5 degrees above the horizontal. The horizontal distance to the net is 7.0 m and the net is 1.0 m high. By how much does the ball clear the net?

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No idea

To find out how much the ball clears the net, we need to calculate the maximum height that the ball reaches (above the net) and then subtract the height of the net.

First, let's find the time it takes for the ball to reach its maximum height using its vertical motion. We can use the equation:

v_fy = v_iy + a*t

where v_fy is the final vertical velocity, v_iy is the initial vertical velocity, a is the acceleration, and t is the time.

The initial vertical velocity can be determined by decomposing the initial velocity of the ball into its x and y components:

v_iy = v_i * sin(theta)

where v_i is the initial velocity (15.0 m/s) and theta is the angle above the horizontal (5.5 degrees).

Using the value of g as the acceleration due to gravity (-9.8 m/s^2), we can set up the equation:

0 = v_iy + (-9.8) * t

Rearranging the equation, we get:

t = v_iy / 9.8

Next, let's find the maximum vertical displacement of the ball using the equation:

Δy = v_iy * t + (1/2) * a * t^2

Substituting the values:

Δy = v_iy * (v_iy / 9.8) + (1/2) * (-9.8) * (v_iy / 9.8)^2

Simplifying the equation will give us:

Δy = (v_iy^2) / (2 * 9.8)

Now, we need to find the height above the net by subtracting the height of the net from the maximum vertical displacement:

Height above net = Δy - net height

Finally, we can substitute the calculated values into the equation to find the height above the net:

Height above net = ((v_i * sin(theta))^2) / (2 * 9.8) - net height

Calculating this equation, we can find the height above the net that the ball clears.