Horizontal Projectile From A Height

For a projectile launched horizontally, the vertical and horizontal motions are completely independent. The vertical motion is free fall from rest, which fixes the time of flight, while the horizontal motion proceeds at constant velocity, which fixes the range. First find the fall time from $h=\frac{1}{2}g{t}^2$: solving $80=\frac{1}{2}(10)t^2$ gives $t^2=16$, so $t=4\text{ s}$. During this same time the package retains its horizontal speed of $15\text{ m/s}$, so the horizontal distance is $x=u_{x}t=15\times 4=60\text{ m}$. The option 45 m wrongly uses a 3 s fall time. The option 80 m mistakenly equates the horizontal range to the drop height. The option 120 m doubles the time of flight by misapplying the free-fall formula. This follows the NCERT independence-of-motions principle for horizontal projectiles. Notably, the time of flight is entirely set by the drop height and would be identical even if the package had simply been dropped from a stationary drone, because the horizontal velocity has no influence on the vertical descent. As a sanity check, the package keeps moving forward with the drone's horizontal velocity throughout the fall, so it lands ahead of and below the release point, and the 60 m range is consistent with a 4 s flight at 15 m/s.