Calculating height in free falling problems

  • Thread starter Thread starter akalei
  • Start date Start date
  • Tags Tags
    Falling Height
Click For Summary
To calculate the time taken for a stone thrown upwards to reach its maximum height, the initial speed is given as 8.0 m/s and the acceleration due to gravity is -10 m/s². The total time from launch to hitting the sea is 3.0 seconds, which includes the ascent and descent phases. The displacement equation s = ut + 1/2at² is used to find the height, but a signage error in the terms needs correction for accurate results. The final calculated height of the cliff is 69 meters. Understanding the direction of velocity and acceleration is crucial in solving free-fall problems.
akalei
Messages
2
Reaction score
0

Homework Statement


A student stands at the edge of a vertical cliff and throws a stone vertically upwards. The stone leaves the student's hand with a speed of v=8.0m/s, the acceleration of free fall is 10m/s2 and all distance measurements are taken from the point where the stone leaves the student's hand. Ignoring air resistance calculate the time taken by the stone to reach its maximum height.

The time between the stone's leaving Antionio's hand and hitting the sea is 3.0s. Determine the height of the cliff.

Homework Equations



unknown

The Attempt at a Solution


s= ut + 1/2at2
s= (8)(3) + 1/2(10)(3)2
s= 69

 
Physics news on Phys.org
If you take positive as up, then negative is down. Which way does vo (u) act? Which way does the acceleration act? Which way is the displacement of the stone to the sea? Don't forget part a.
 
i think i understand what your saying but i don't understand how to use the information to solve the problem
 
For the equation you wrote for displacement, you have a signage error in one of the terms on the right side of the equation. Can you spot it??
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
1
Views
2K
Replies
23
Views
3K
  • · Replies 3 ·
Replies
3
Views
6K
Replies
2
Views
2K
Replies
6
Views
2K
  • · Replies 13 ·
Replies
13
Views
11K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 6 ·
Replies
6
Views
2K