Projectile thrown horizontally off a cliff

  • Thread starter jeff12
  • Start date
  • Tags
    Projectile
In summary, the conversation is about a ball being thrown horizontally from a height of 20m and hitting the ground at a speed that is three times its initial speed. The participants discuss the time of flight, the acceleration of the ball, and the components of velocity in order to calculate the initial speed. They also consider the effects of gravity and the direction of the velocity. Ultimately, they come to the conclusion that the final horizontal component of velocity is the same as the initial, and the speed of the ball can be calculated using the components of velocity.
  • #1
jeff12
40
2
Member warned to retain and use the formatting template
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?
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
I am confused now but I was told that horizontally an object has no acceleration?
 
  • #6
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.
 
  • #7
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.
 
  • #8
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?
 
  • #9
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?
 
  • Like
Likes CWatters
  • #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

[tex]\Delta \ x = \ v_0 t + \frac{1}{2} \alpha t^2[/tex]

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?
 

1. What is a projectile thrown horizontally off a cliff?

A projectile thrown horizontally off a cliff is an object that is launched horizontally from the edge of a cliff with an initial velocity. The object follows a parabolic path due to the force of gravity, eventually landing on the ground below.

2. How does the initial velocity affect the trajectory of a projectile thrown horizontally off a cliff?

The initial velocity of a projectile thrown horizontally off a cliff determines the speed at which the object travels horizontally. However, the vertical component of the velocity remains constant due to the force of gravity. This results in a curved trajectory for the object.

3. What is the formula for calculating the horizontal displacement of a projectile thrown horizontally off a cliff?

The formula for calculating the horizontal displacement of a projectile thrown horizontally off a cliff is d = v * t, where d is the horizontal displacement, v is the initial velocity, and t is the time elapsed. This assumes that there is no air resistance acting on the object.

4. How does the height of the cliff affect the trajectory of a projectile thrown horizontally off a cliff?

The height of the cliff does not affect the horizontal displacement of the projectile. However, it does affect the time it takes for the object to reach the ground. The higher the cliff, the longer the time of flight for the object.

5. What factors can affect the trajectory of a projectile thrown horizontally off a cliff?

The trajectory of a projectile thrown horizontally off a cliff can be affected by factors such as air resistance, wind, and the angle of the initial velocity. These factors can alter the speed and direction of the object, resulting in a different trajectory. Additionally, the mass and shape of the object can also affect its trajectory.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
206
  • Introductory Physics Homework Help
Replies
4
Views
877
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
2
Replies
38
Views
1K
  • Introductory Physics Homework Help
Replies
34
Views
3K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
827
  • Introductory Physics Homework Help
Replies
19
Views
2K
Back
Top