Questions: A projectile is launched horizontally off a cliff of known height h with speed v0 and reaches the ground a time t after being launched. Another identical projectile is launched horizontally off the same cliff with speed 2v0. How long does it take the second projectile to reach the ground? A t/2 (B) t (C) 2t (D) 4t

A projectile is launched horizontally off a cliff of known height h with speed v0 and reaches the ground a time t after being launched. Another identical projectile is launched horizontally off the same cliff with speed 2v0. How long does it take the second projectile to reach the ground?

A t/2
(B) t
(C) 2t
(D) 4t

Transcript text: A projectile is launched horizontally off a cliff of known height $h$ with speed $v_{0}$ and reaches the ground a time $t$ after being launched. Another identical projectile is launched horizontally off the same cliff with speed $2 v_{0}$. How long does it take the second projectile to reach the ground? A $\frac{t}{2}$ (B) $t$ (C) $2 t$ (D) $4 t$

Solution

Solution Steps

Step 1: Understanding the Problem

We need to determine the time it takes for a second projectile, launched horizontally with twice the speed of the first, to reach the ground from the same height.

Step 2: Analyzing the First Projectile

The first projectile is launched horizontally from a height $ h $ and reaches the ground in time $ t $. The time to reach the ground depends only on the vertical motion, which is governed by the equation: \[ h = \frac{1}{2} g t^2 \] where $ g $ is the acceleration due to gravity.

Step 3: Analyzing the Second Projectile

The second projectile is launched horizontally from the same height $ h $ but with twice the horizontal speed. The vertical motion is still governed by the same equation: \[ h = \frac{1}{2} g t_2^2 \] where $ t_2 $ is the time it takes for the second projectile to reach the ground.

Step 4: Comparing the Times

Since the vertical motion is independent of the horizontal speed, the time $ t_2 $ for the second projectile to reach the ground is the same as the time $ t $ for the first projectile: \[ t_2 = t \]