Question

.51 q: projectile a is launched horizontally at a speed of 20 meters per second from the top of a cliff and strikes a level surface below, 3.0 seconds later. projectile b is launched horizontally from the same location at a speed of 30 meters per second. the time it takes projectile b to reach the level surface is (a) 4.5 s (b) 2.0 s (c) 3.0 s (d) 10 s

Explanation:

Step1: Analyze vertical motion

The vertical - motion of a horizontally launched projectile is a free - fall motion. The initial vertical velocity \(v_{0y}=0\ m/s\), and the vertical displacement \(y = v_{0y}t+\frac{1}{2}gt^{2}=\frac{1}{2}gt^{2}\) (where \(g\) is the acceleration due to gravity and \(t\) is the time of flight). The time of flight of a horizontally launched projectile depends only on the vertical displacement and the acceleration due to gravity.

Step2: Compare projectiles A and B

Since projectile A and projectile B are launched from the same vertical height (same cliff) and the vertical motion of both projectiles is governed by the same equation \(y=\frac{1}{2}gt^{2}\) (and the initial vertical velocity for both is \(0\ m/s\)), the time of flight is the same for both projectiles regardless of their horizontal launch speeds. Projectile A takes \(t_A = 3.0\ s\) to reach the level surface. So, projectile B will also take \(t_B=3.0\ s\) to reach the level surface.

Answer:

C. 3.0 s