Question

Mark is in a deep hole looking for treasure. He is standing 8 feet below the surface. He throws an old watch he found with an initial upward velocity of 23 ft/sec. How long until it lands outside the hole, having gone up and back down?

Answer 1

To find the time until the watch lands outside the hole, use the equations of motion for a projectile in free fall.

Step-by-step explanation:

To find the time until the watch lands outside the hole, we can use the equation of motion for a projectile in free fall. Since the watch is thrown upward and falls back down, we can consider it as two separate motions. Here's how we can find the time:

1. First, find the time it takes for the watch to reach maximum height. Use the equation: t = v / g, where t is the time, v is the initial upwards velocity, and g is the acceleration due to gravity (approximately 32 ft/s^2).
2. Next, find the total time for the watch to land outside the hole. Since the watch takes the same amount of time to fall back down as it did to reach maximum height, simply multiply the time from step 1 by 2.

The total time until the watch lands outside the hole will be the answer.

Answer 2

He trows a watch with an initial upward velocity: v o = 23 ft/s.
v = v o - g · t
At the highest point:
0 = 23 - 9.81 · t
9.81 t = 23
t = 23 : 9.81
t = 2.34 s
h = v o · t - g t² / 2
h = 23 · 2.34 - 9.81 · 2.34² / 2
h = 53.92 - 26.86 = 26.86 ft
26.86 ft - 8 ft = 18.86 ft
18.86 = g · t² / 2
t² = 18.86 · 2 : 9.81
t = 1.96 s
t ( total ) = 2.34 s + 1.96 s = 4.3 s
Answer: The watch will land outside the hole after 4.3 seconds.