(a) The ball falls short of clearing the crossbar by 3.05 m (negative value indicates falling short).
(b) The ball approaches the crossbar while falling since it doesn't reach a height greater than the crossbar's height during its trajectory.
To solve this problem, we'll analyze the vertical motion of the ball.
(a) To find how much the ball clears or falls short of clearing the crossbar vertically, we need to calculate the maximum height reached by the ball.
The initial velocity (V0) of the ball is 23.2 m/s, and the launch angle (θ) is 52.0° above the horizontal.
The vertical component of velocity (Vy) at the highest point of the trajectory is zero since the ball momentarily stops before falling back down.
To find the time taken to reach the highest point, we can use the equation:
Vy = V0 * sin(θ)
0 = 23.2 m/s * sin(52.0°)
Solving for sin(52.0°), we find:
sin(52.0°) ≈ 0.7880
Dividing both sides by 23.2 m/s, we get:
0.7880 = sin(52.0°)
Taking the inverse sine, we find:
52.0° ≈ arcsin(0.7880)
Using a calculator, we find:
52.0° ≈ 56.43°
Now we can calculate the time (t) it takes to reach the highest point using the equation:
t = (2 * Vy) / g
Since Vy = 0, we have:
t = 0
This means that the ball reaches its maximum height instantaneously and starts falling immediately. Therefore, the ball does not clear the crossbar.
To find how much the ball falls short of clearing the crossbar vertically, we can calculate the height of the ball at a horizontal distance of 36.0 m.
Using the equation for vertical displacement, we have:
Δy = V0y * t + (1/2) * g * [tex]t^2[/tex]
Plugging in the known values:
Δy = 0 * t + (1/2) * (-9.8 [tex]m/s^2[/tex]) * ([tex]t^2[/tex])
Since t = 0, the equation simplifies to:
Δy = 0
Therefore, the ball falls short of clearing the crossbar by 3.05 m vertically.
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