Solving for Displacement when Jamilla Throws a Stone

• sciencegeek101
In summary: Note:- If you also take the direction in account then the value of y will be negative. So, you can write the final answer as:-The stone was 20 meters above the water's surface when Jamilla threw it.In summary, the stone that Jamilla threw horizontally off a pier into the ocean at a velocity of 20 m/s was 20 meters above the water's surface when she threw it. This was determined by calculating the stone's vertical displacement using the formula Δd= (Vf2-Vi2) / 2a, where Vf represents the final velocity, Vi represents the initial velocity, and a represents the acceleration due to gravity. The horizontal distance traveled by the stone was not taken into account
sciencegeek101
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!

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:

Your calculations are correct! To solve for displacement, we use the equation Δd= (Vf2-Vi2) / 2a, where Vf is the final velocity, Vi is the initial velocity, and a is acceleration. In this case, the acceleration is due to gravity, which is 9.8m/s2. By plugging in the given values, we can solve for Δd, which represents the displacement or the height of the stone above the water's surface when Jamilla threw it. Your final answer of 20.4m is correct. Great job!

1. How do you calculate displacement when Jamilla throws a stone?

The formula for calculating displacement is displacement = final position - initial position. In this case, the final position would be where the stone lands and the initial position would be where Jamilla released the stone from her hand.

2. Does the mass of the stone affect the displacement?

No, the mass of the stone does not affect the displacement. Displacement is solely determined by the distance between the initial and final positions.

3. What units should displacement be measured in?

Displacement is typically measured in meters (m) or feet (ft). However, any unit of length can be used as long as it is consistent throughout the calculation.

4. How does the angle of Jamilla's throw affect the displacement?

The angle of Jamilla's throw affects the displacement by changing the direction of the stone's motion. The magnitude of the displacement will remain the same, but the direction will change based on the angle of the throw.

5. Can displacement be negative?

Yes, displacement can be negative. This indicates that the final position is behind the initial position, or that the object has moved in the opposite direction of its initial position.

• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
3K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
4K
• Introductory Physics Homework Help
Replies
15
Views
6K
• Introductory Physics Homework Help
Replies
7
Views
3K
• Introductory Physics Homework Help
Replies
10
Views
3K
• Introductory Physics Homework Help
Replies
13
Views
4K