Mechanics · Projectile Motion

Track the trajectory through P and locate the maximum height

A projectile launched from O must pass through P at horizontal distance 5 m and height 1 m. The trajectory equation fixes the required speed for each launch angle and also tells whether the highest point comes before or after P.

2026 Paper 1 Q6 · projectile through P(5, 1)
Statement check
Launch angle 45°
Trajectory condition
The projectile must satisfy y = x tanθ − gx²/(2v²cos²θ) at x = 5 m and y = 1 m.
Relation1 = 5tanθ − 25g/(2v²cos²θ)
VerdictA and B are correct
Projectile pathtrajectory satisfying P(5, 1)
Apexmaximum height position
Point Pgiven point at 5 m, 1 m

Derivation path

01 / TRAJECTORY EQUATION

Substitute P(5, 1)

The projectile equation gives the speed needed for any allowed launch angle.

v² = 25g / [2cos²θ(5tanθ − 1)]
02 / θ = 45°

Statement A becomes true

With tan45° = 1 and cos²45° = 1/2, the speed is fixed as stated.

v² = 25g/4, so v = 5√g/2
03 / APEX LOCATION

Maximum height occurs before P

For θ = 45°, the apex lies at x = v²sinθcosθ/g = 25/8 m, which is less than 5 m.

xmax = 25/8 m < 5 m
Correct statements
Statement A is true from the trajectory equation at θ = 45°.
Statement B is true because the apex lies before P for θ = 45°.
Correct answer: A and B