Projectile thrown horizontally off a cliff

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Homework Help Overview

The problem involves a ball thrown horizontally from a height of 20 meters, which impacts the ground with a speed that is three times its initial speed. The discussion centers on understanding the components of motion, particularly the effects of gravity on vertical and horizontal acceleration.

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  • Mixed

Approaches and Questions Raised

  • Participants explore the time of flight and question the nature of acceleration during horizontal motion. There are discussions about the relationship between vertical and horizontal components of velocity and how to express these in terms of initial speed.

Discussion Status

Participants have provided insights into the nature of projectile motion, emphasizing the role of gravity and the independence of horizontal motion from vertical acceleration. There is ongoing exploration of how to relate the final speeds in both components and the implications of these relationships.

Contextual Notes

Some participants express confusion regarding the acceleration of the ball once it is in flight, particularly in relation to its horizontal motion. There are references to differing methods taught by instructors and the need to clarify the relationship between initial and final speeds in the context of projectile motion.

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A ball is thrown horizontally from a height of 20m and hits the ground with a speed that is three times its initial speed. What is the initial speed?d=vit+1/2at^2I found the time of flight vertically first, which is 2.02 sec. And then I am trying to find the initial speed. Even if it is thrown horizontally off a cliff, is the acceleration still 0?
 
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jeff12 said:
I found the time of flight vertically first, which is 2.02 sec. And then I am trying to find the initial speed. Even if it is thrown horizontally off a cliff, is the acceleration still 0?
Did you mean to say 0 here? Once the ball leaves your hand, gravity takes over. So, what is the acceleration (magnitude and direction) while the ball is in flight?

Your value for the time of flight looks correct.
 
Doesnt it depend? Horizontally it does not have a acceleration. But does it apply when it is thrown off a cliff? Because it is thrown off the cliff horizontally not straight down.
 
No matter how you throw the ball, once it's in flight the only thing "acting on it" is gravity (neglecting air resistance, of course). The acceleration caused by gravity is g downward.
 
I am confused now but I was told that horizontally an object has no acceleration?
 
Acceleration is a vector quantity. For a projectile, the horizontal component of the acceleration is zero. But the vertical component is 9.8 m/s2 downward.
 
Can I solve this problem horizontally? I tried it but it seems a little off.

Vf=Vi+at
A would be canceled because of horizontally
Vf=Vi?
I looked in the answer key and they did it differently but my professor taught us this way to find vertically and horizontally. I don't know if it applies to this problem though.
 
Yes, you should think about the horizontal and vertical components of the motion.

Suppose Vo is the initial speed that the object is thrown horizontally. Since the horizontal component of acceleration is zero, what can you say about the value of the horizontal component of velocity just as the ball reaches the ground?
 
TSny said:
Yes, you should think about the horizontal and vertical components of the motion.

Suppose Vo is the initial speed that the object is thrown horizontally. Since the horizontal component of acceleration is zero, what can you say about the value of the horizontal component of velocity just as the ball reaches the ground?

It's the same as the initial?
 
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  • #10
Yes. Now think about the vertical component of motion. What is the initial value of the vertical component of velocity? Can you see a way to get the final vertical component of velocity using the height that the ball was thrown from?
 
  • #11
TSny said:
Yes. Now think about the vertical component of motion. What is the initial value of the vertical component of velocity? Can you see a way to get the final vertical component of velocity using the height that the ball was thrown from?

I got Vf for the vertical component as 19.8. How is this related to the Vf of the horizontal?
 
  • #12
Your post #9 is correct. The horizontal component is unchanged.
 
  • #13
Bump
 
  • #14
jeff12 said:
I got Vf for the vertical component as 19.8. How is this related to the Vf of the horizontal?
I'm not sure I understand your question here. But, how do you construct the speed of a particle from it's horizontal and vertical components?
 
  • #15
TSny said:
I'm not sure I understand your question here. But, how do you construct the speed of a particle from it's horizontal and vertical components?

You would just use

\Delta \ x = \ v_0 t + \frac{1}{2} \alpha t^2

for both horizontal and vertical
 
  • #16
You just add (vector add) the horizontal and vertical components to get the impact speed.
 
  • #17
Let's make sure we are together so far. Let Vo be the initial speed of the ball.

When the ball reaches the ground, how would you express the horizontal and vertical components of the velocity?
 
  • #18
Vertically the final velocity would be 39.2. I don't know how to find it horizontally.
 
  • #19
jeff12 said:
Vertically the final velocity would be 39.2. I don't know how to find it horizontally.
In post #11 you stated that the vertical component would be 19.8 m/s. And in post #9 you gave the correct answer for the horizontal component of the final velocity.
 
  • #20
TSny said:
In post #11 you stated that the vertical component would be 19.8 m/s. And in post #9 you gave the correct answer for the horizontal component of the final velocity.

The 19.8 is the final velocity of the vertical.

Vf=Vi only in the horizontal component not the vertical component.
 
  • #21
Yes, the final vertical component of velocity is 19.8 m/s2 downward. That is Vfy= -19.8 m/s2 (if we take the positive y direction to be upward).

Suppose you let the symbol Vo stand for the initial speed of the particle when it was thrown horizontally. You are asked to find the value of Vo such that the final speed is 3 times Vo. It will help if you can answer the following.

(1) How does the final horizontal component of velocity compare to Vo? That is, can you express Vfx in terms of Vo?

(2) How do you calculate the speed of a particle if you know the values of the horizontal and vertical components of velocity, Vx and Vy?
 
  • #22
TSny said:
Yes, the final vertical component of velocity is 19.8 m/s2 downward. That is Vfy= -19.8 m/s2 (if we take the positive y direction to be upward).

Suppose you let the symbol Vo stand for the initial speed of the particle when it was thrown horizontally. You are asked to find the value of Vo such that the final speed is 3 times Vo. It will help if you can answer the following.

(1) How does the final horizontal component of velocity compare to Vo? That is, can you express Vfx in terms of Vo?

(2) How do you calculate the speed of a particle if you know the values of the horizontal and vertical components of velocity, Vx and Vy?

1) Vo=Vf/3
2) This why I am asking...is there some sort of physics rule?
 
  • #23
jeff12 said:
1) Vo=Vf/3
It's important to see what's going on with just the horizontal component of the velocity. How would you express the initial horizontal component of velocity, Vix, in terms of Vo? How would you express the final horizontal component of velocity, Vix, in terms of Vo?

2) This why I am asking...is there some sort of physics rule?
Speed is defined to be the magnitude of the velocity vector. How do you calculate the magnitude of a vector using the components of the vector?
 

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