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A projectile thrown so that it passes through a point 1 m high and 5 m away: its launch speed and where it reaches its highest point

A launcher at O throws a ball through a ring P, 1 m high and 5 m away. At 45 degrees it needs v = 5 sqrt(g)/2 and peaks at 3.125 m, before P; at 30 degrees it peaks at 3.825 m, also before P; along tan inverse 1/5 even 125 sqrt(g) passes below P - answer (A), (B).

Sub-topics covered

• Equation of trajectory y = x tan th - g x^2/(2 v^2 cos^2 th)
• Speed needed to pass a given point at a given angle
• Highest point at x = v^2 sin 2th / 2g (half the level range)
• Aiming along the line of sight needs infinite speed: gravity pulls every real path below it

What it teaches

• Through P (5 m, 1 m): v^2 = 25 g / (2 cos^2 th (5 tan th - 1))
• At 45 deg v = 5 sqrt(g)/2 and the top is at x = 25/8 = 3.125 m, before P - statements (A) and (B)
• At 30 deg v^2 = 8.83 g and the top is at x = 3.825 m, still before P - statement (C) false
• tan th = 1/5 is the slope of OP itself, so 5 tan th - 1 = 0 and v is unbounded; 125 sqrt(g) passes 0.83 mm below P - statement (D) false
• The top lies beyond P only for shallow throws, 11.3 deg < th < 21.8 deg

Simulation ID

• ADV-2026-P1-PHY-Q06 · revision 1

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