Solving for Displacement when Jamilla Throws a Stone

  • Thread starter Thread starter sciencegeek101
  • Start date Start date
  • Tags Tags
    Displacement
Click For Summary

Homework Help Overview

The problem involves determining the height from which a stone is thrown horizontally off a pier, given its horizontal velocity and the distance it travels before hitting the water. The subject area pertains to projectile motion and kinematics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the separation of horizontal and vertical motion components, emphasizing the lack of horizontal acceleration and the need to calculate time of flight based on horizontal distance and speed. Some participants suggest using equations for projectile motion to relate the horizontal distance to the vertical height.

Discussion Status

There is an ongoing exploration of the problem with various interpretations of the motion involved. Some participants have provided guidance on breaking down the motion into components, while others have pointed out potential misunderstandings in the original approach. No consensus has been reached yet.

Contextual Notes

Participants are navigating the complexities of projectile motion, particularly the relationship between horizontal and vertical components, and the implications of initial conditions such as velocity and acceleration. There is also a focus on ensuring the correct application of kinematic equations.

sciencegeek101
Messages
8
Reaction score
0
Jamilla throws a stone horizontally off of a pier into the ocean at a velocity of 20 m/s. If the stone is 40.0 m from the edge of the pier when it hits the water, how high above the water's surface was the stone when Jamilla threw it?

Vf= 20m/s
Vi= 0m/s
a=9.8m/s2
Δd= 40m
Δd=?

Vf2=Vi2+2aΔd

Vf=velocity final
Vi= velocity initial
a= acceleration
Δd= displancement

(rearrange the equation)
Δd= (Vf2-Vi2) / 2a
Δd= (20^2-o^2) / 2x9.8
Δd= 400/19.6
Δd= 20.4m

∴ the stone was 20.4m above the water surface when Jamilla threw it

Not sure if i am doing this correctly?
Much thanks in advance to those who help me out!
 
Physics news on Phys.org
Be careful - the initial velocity is horizontal, but there is no horizontal acceleration. You have that the initial vertical velocity component is 0 m/s, the vertical acceleration is 9.8 m/s^2 downwards, and you don't know the final vertical velocity. You know the horizontal distance traveled is 40m, and the horizontal speed is constant at 20 m/s since there's no horizontal acceleration, so you can find the time of flight using speed = distance / time.

Then you'll have enough information to calculate the height using the vertical information...
 
What Naz93 said. Break it down into horizontal and vertical components and solve in two stages. You appear to have them a bit mixed up.
 
You are approaching the wrong way. The stone travels both in horizontal as well as vertical direction. And motion in two different directions has nothing to do with each other. Here, the acceleration due to gravity acts only in the y-direction. So, the stone accelerates downward. On the other hand, there is no acceleration in x-direction. Hence, the stone undergoes uniform motion in x-direction. It is called a projectile motion:-

x= v(i,x)* time v(i,x): initial velocity in x-direction
y= 1/2* g * time^2 g:acceleration due to gravity note: v(i,y)=0

Replacing time from both the equations:-

y=(g * x^2)/(2*(v(i,x))^2)

Putting values:-

y=(10* 40^2)/(2* 20^2)= 20 m
 
Last edited:

Similar threads

Replies
5
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
3
Views
3K
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
7
Views
4K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 15 ·
Replies
15
Views
6K