Velocities with a resulting horizontal component

Click For Summary
To determine the speed of the pebbles when they hit Juliet's window, it is essential to analyze the projectile motion, which consists of both vertical and horizontal components. The vertical component must reach a height of 8 meters, with the vertical velocity being zero at the apex of the throw. By using the equations of motion, the time taken to reach this height can be calculated, allowing for the horizontal distance of 9 meters to be covered in that same time. The horizontal speed can then be derived from the initial velocity and angle of projection. Understanding these components is crucial for solving the problem accurately.
Surgikill117
Messages
12
Reaction score
0

Homework Statement


Romeo is chucking pebbles gently up to Juliet's window, and he wants the pebbles to hit the window with a horizontal component of velocity. He is standing at the edge of a rose garden 8m below her window and 9m from the base of the wall. How fast are the pebbles going when they hit her window.


Homework Equations


d=.5at^2
t=d/v
t=√2d/a

The Attempt at a Solution


I don't know how to find the speed when it has only a horizontal component of speed. Am I supposed to find how fast he should throw it so its apex is at her window?
 
Physics news on Phys.org
A velocity has both vertical and horizontal components of velocity but the acceleration due to gravity applies only to the vertical component. Determine what the initial vertical velocity must be so that it is 0 at a height of 8m. Use that to determine the time it takes the pebble to go that high. Determine the horizontal speed speed so that the pebble will go 9 m in that time.
 
Surgikill117 said:

Homework Statement


Romeo is chucking pebbles gently up to Juliet's window, and he wants the pebbles to hit the window with a horizontal component of velocity. He is standing at the edge of a rose garden 8m below her window and 9m from the base of the wall. How fast are the pebbles going when they hit her window.


Homework Equations


d=.5at^2
t=d/v
t=√2d/a

The Attempt at a Solution


I don't know how to find the speed when it has only a horizontal component of speed. Am I supposed to find how fast he should throw it so its apex is at her window?
Yes. At the apex of the arc, the derivative of the vertical displacement with respect to the horizontal displacement is 0. In non-mathy terms, this means that the pebble is moving horizontally at the apex of the arc.
 
The main idea is to remember that you can decompose this type of motion (known as projectile motion) into two parts; the motion in the vertical direction and the motion in the horizontal direction. If you take the origin of your coordinate system to be ground (assuming falsely that Romeo has zero height, or is otherwise in a hole in the ground such that he can throw the pebble at ground level) you will find that the equation of motion in the vertical component will be:
$$ y(t) = \frac{-1}{2}gt^2 + V_0\sin(\theta)t $$

Where V_0\sin(\theta) is the vertical component of the initial velocity. The horizontal motion is given by:
$$ x(t) = V_0\cos(\theta)t $$
Where V_0\cos(\theta) is the horizontal component of the initial velocity. Now, we want that the vertical component of the velocity to be zero when the height above the ground is 8 m. From the vertical motion equation we can find the general expresion for the vertical velocity:
$$ V_y(t) = -gt + V_0\sin(\theta) $$
In order to find the time it takes the pebble to reach its apex (where its vertical velocity is zero) set the previous equation to zero and solve for the time t. Substitute this value into the motion equation and set it equal to 8 (Since at that time the height of the pebble will be 8). This will allow you to solve for an expression involving the initial velocity and the initial angle \theta. Now, you know that the horizontal motion will be 9 m, so use that to solve for V_0 and \theta. I'll be happy to provide any more detail as needed.
 
Last edited:
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
20
Views
6K
Replies
13
Views
3K
  • · Replies 2 ·
Replies
2
Views
7K
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 2 ·
Replies
2
Views
9K
  • · Replies 8 ·
Replies
8
Views
6K
  • · Replies 3 ·
Replies
3
Views
6K