Projectile Launch Over a Fence
A ball is launched from ground level at 18 m/s at 40° above the horizontal toward a 2.4 m high fence whose base is 22 m from the launch point. Air resistance is negligible. A student claims the ball will clear the fence because the maximum height of the trajectory (about 6.8 m) is well above 2.4 m. The launch point and the fence base are on level ground, and the fence is treated as a vertical line at horizontal distance 22 m. The student has not considered where along the path the ball is when it reaches the fence.
1. Determine whether the ball clears the fence, showing the horizontal and vertical components separately.
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Horizontal component: v_x = 18 cos40° ≈ 13.79 m/s. Vertical component: v_y = 18 sin40° ≈ 11.57 m/s.
Time to reach the fence's horizontal position: t = 22 / 13.79 ≈ 1.60 s.
Height at that time: y = v_y t − ½ g t² = 11.57(1.60) − 4.9(1.60)² ≈ 18.5 − 12.5 ≈ 6.0 m.
Since 6.0 m > 2.4 m, the ball clears the fence by about 3.6 m. The student's conclusion is correct, though the stated reason (peak height) is not the relevant test; the relevant test is height at the fence's horizontal position.
2. The student then argues that because peak height is 6.8 m, the ball would clear a fence of any height up to 6.8 m placed anywhere. Identify the flaw and state the condition under which this reasoning fails.
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Peak height only applies at the single horizontal position where v_y = 0, at t = v_y/g ≈ 1.18 s, i.e. x ≈ 16.3 m. At other horizontal positions the ball is lower.
The reasoning fails whenever the fence is not at x ≈ 16.3 m. For a fence at a larger x, the ball is descending and its height is less than the peak; for a fence closer than 16.3 m the ball is still rising and is also below the peak. So a fence of height between the height-at-fence and 6.8 m would be cleared only if placed near the peak.
Correct general condition: the ball clears a fence of height h at horizontal distance d if y(d) > h, where y(d) = d tanθ − g d²/(2 v² cos²θ).