Projectile Range And Two Angles

Question

A ball thrown from level ground at a fixed speed lands a certain horizontal range away when launched at 25 degrees above the horizontal. At which other launch angle, using the same speed, will the ball achieve exactly the same horizontal range?

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The horizontal range of a projectile launched over level ground at speed $u$ and angle $	heta$ is $R = \frac{u^2\sin 2	heta}{g}$. Because the range depends on $\sin 2	heta$, two different launch angles can produce the same range whenever their doubled angles have the same sine. The sine function satisfies $\sin 2	heta = \sin(180^\circ - 2	heta)$, so two angles $	heta_1$ and $	heta_2$ give equal range when $2	heta_2 = 180^\circ - 2	heta_1$, that is when $	heta_1 + 	heta_2 = 90^\circ$. Thus the complementary angle to $25^\circ$ is $90^\circ - 25^\circ = 65^\circ$, which yields an identical range at the same speed. The option 75 degrees is wrong because it does not sum to 90 degrees with 25. The option 55 degrees likewise fails the complementary condition. The option 45 degrees is the single angle of maximum range, not a partner to 25 degrees. This is the standard JEE result on complementary projectile angles. As a check, the pair 25 and 65 degrees straddles the 45-degree maximum symmetrically, and a steeper launch trades higher flight time for lower horizontal speed to recover the same range.

Projectile Range And Two Angles

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