Solving Physics Problems: Dropped Stones, Thrown Balls, and Falling Tiles

[SomeHustler]

1. A spelunker (cave explorer) drops a stone from rest into a hole. The speed of sound is 343 m/s in air, and the sound of the stone striking the bottom is heard 1.64 s after the stone is dropped. How deep is the hole?

2. A ball is thrown upward from the top of a 25.0-m-tall building. The ball's initial speed is 12 m/s. At the same instant, a person is running on the ground at a distance of 32.6 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?

3. A roof tile falls from rest from the top of a building. An observer inside the building notices that it takes 0.18 s for the tile to pass her window, whose height is 1.8 m. How far above the top of this window is the roof?

Homework Equations

Acceleratior = (Vf + Vi) / time displacement

Vector average = (Vf + Vi) / 2

Vector Average = distance displacement / time displacement

The Attempt at a Solution

No clue, if someone could help me that would be great. I can't do either of those 3 for some reason.

 

[learningphysics]

Give these a shot... all of them involve acceleration due to gravity. Have you studied equations for constant acceleration?

 

[m2003]

I can provide some guidance and steps to solve these physics problems:

1. To solve the first problem, we can use the equation for the speed of sound: v = d/t, where v is the speed of sound, d is the distance the sound travels, and t is the time it takes for the sound to travel. We know the speed of sound (343 m/s) and the time it takes for the sound to reach the observer (1.64 s). Since the sound has to travel down and then back up, we can divide the time by 2 to get the time it takes for the stone to reach the bottom of the hole (0.82 s). Using this time, we can rearrange the equation to solve for the distance, d = v*t. Plugging in the values, we get d = 343 m/s * 0.82 s = 281.26 m. Therefore, the depth of the hole is approximately 281 meters.

2. For the second problem, we can use the equations of motion for a ball thrown upward. The ball's initial velocity is 12 m/s and it is thrown from a height of 25.0 m. We need to find the time it takes for the ball to reach the ground. Using the equation h = Vi*t + 1/2*a*t^2, where h is the height, Vi is the initial velocity, a is the acceleration due to gravity (9.8 m/s^2), and t is the time, we can solve for t. Plugging in the values, we get 25.0 m = 12 m/s * t + 1/2 * (-9.8 m/s^2) * t^2. This is a quadratic equation that we can solve to find t. We get two solutions, t = 2.06 s and t = -1.89 s. Since we are only interested in the positive value, we can discard the negative solution. Therefore, it takes approximately 2.06 seconds for the ball to reach the ground. Now, we can use the equation v = Vi + a*t to find the final velocity of the ball just before it reaches the ground. Plugging in the values, we get v = 12 m/s + (-9.8 m/s^2) * 2.06 s = -8.12 m/s. This means the ball is