What is the speed of the ball when it hits the ground?

In summary: So you do not need to know them separately.In summary, the speed of the ball when it hits the ground below the cliff is 24.4 m/s. This can be calculated using the conservation of energy equation, which takes into account the initial velocity and the change in potential energy due to the drop.
  • #1
JaZZyCooL
7
0

Homework Statement


A ball is thrown from the edge of a 10.0 m high cliff with a speed of 20.0 m/s at an angle of 30 degrees below horizontal. What is the speed of the ball when its hits the ground below the cliff?

Options :
1) 26.3 m/s
2) 28.1 m/s
3) 31.4 m/s
4) 24.4 m/s
5) 29.5 m/s

I am new to the forum so bare with me I don't know how the forum works. I have attempted a solution for the problem. I would appreciate if you could just tell me the equation to work with.

Homework Equations



This is the equation I tried

v = sqrt(2*g*h)

The Attempt at a Solution



v = sqrt(2*9.8*10)
v = 14 m/s

The answer is not even in the options. Clearly I am doing something wrong actually way off. May be I am mixing wrong concept here.

I would appreciate explanation with equation.

Thank you.
 
Physics news on Phys.org
  • #2
JaZZyCooL said:

Homework Statement


A ball is thrown from the edge of a 10.0 m high cliff with a speed of 20.0 m/s at an angle of 30 degrees below horizontal. What is the speed of the ball when its hits the ground below the cliff?

Options :
1) 26.3 m/s
2) 28.1 m/s
3) 31.4 m/s
4) 24.4 m/s
5) 29.5 m/s

I am new to the forum so bare with me I don't know how the forum works. I have attempted a solution for the problem. I would appreciate if you could just tell me the equation to work with.

Homework Equations



This is the equation I tried

v = sqrt(2*g*h)

The Attempt at a Solution



v = sqrt(2*9.8*10)
v = 14 m/s

The answer is not even in the options. Clearly I am doing something wrong actually way off. May be I am mixing wrong concept here.

I would appreciate explanation with equation.

Thank you.

Welcome to the PF.

JaZZyCooL said:
v = sqrt(2*g*h)

That would be the speed if the ball were dropped from the height h. But it was thrown, so there will be some sort of addition of speeds...:smile:
 
  • Like
Likes topsquark
  • #3
JaZZyCooL said:
A ball is thrown from the edge of a 10.0 m high cliff with a speed of 20.0 m/s at an angle of 30 degrees below horizontal. What is the speed of the ball when its hits the ground below the cliff?
JaZZyCooL said:
v = sqrt(2*9.8*10)
v = 14 m/s
your ball also travels horizontally not just vertically try adding that to your equation
 
  • Like
Likes topsquark
  • #4
So is it v=sqrt(2*9.8*10) +20
so V = 14 m/s + 20 m/s

v = 24 m/s?
 
  • #5
JaZZyCooL said:
So is it v=sqrt(2*9.8*10) +20
so V = 14 m/s + 20 m/s

v = 24 m/s?

14+20 is not 24...

Besides, did you see the hint by Matejxx1? The velocity of the ball has horizontal and vertical components. Gravity affects the vertical component...
 
  • Like
Likes topsquark
  • #6
v = sqrt (Vi^2 + 2*g*h)
v = sqrt( 20^2 + 2*9.8*10)

v = 24.4 m/s ?
 
  • #7
JaZZyCooL said:
v = sqrt (Vi^2 + 2*g*h)
v = sqrt( 20^2 + 2*9.8*10)

v = 24.4 m/s ?

That looks good. You could have also solved it by finding the initial Vx and Vy components, then added in the extra Vy component due to the acceleration of gravity over the 10m drop, and then taken the SQRT of the squares of the x & y velocity components. Maybe try that method to be sure you get the same answer. :smile:
 
  • Like
Likes topsquark
  • #8
I calculated the initial x direction speed (17.4 m/s) and y direction (up & down) 10 m/s using simple trigonometry. The x direction speed does not change (assumes no air friction) and the y direction changes to 17.2 m/s calculated by adding the initial velocity (10 m/s) with the added velocity by dropping 10 m (t = 0.735 s drop). The square root of the sum of the squares yields the 24.4 m/s answer.
 
  • #9
phsicsgeek said:
I calculated the initial x direction speed (17.4 m/s) and y direction (up & down) 10 m/s using simple trigonometry. The x direction speed does not change (assumes no air friction) and the y direction changes to 17.2 m/s calculated by adding the initial velocity (10 m/s) with the added velocity by dropping 10 m (t = 0.735 s drop). The square root of the sum of the squares yields the 24.4 m/s answer.
You are responding to a thread that is seven years old.

The approach of separating the velocities is workable. But there is a shortcut. Conservation of energy. Mass is irrelevant. So pick a mass and call it ##m##. You know starting speed. So you know starting kinetic energy. You know potential energy. So you know final kinetic energy. So you know final speed. No trig involved. [Work algebraicly. Derive formula. Evaluate formula. That way the mass ##m## drops out cleanly].

It'll still come out as the square root of a sum involving squares. You just do not have to separate out the squared initial horizontal velocity component from the squared initial vertical velocity component. Pythagoras already tells you the sum of those two squares.
 
Last edited:
  • Like
Likes Curiosity_0

Related to What is the speed of the ball when it hits the ground?

1. What is the speed of the ball when it hits the ground?

The speed of the ball when it hits the ground depends on several factors, such as the initial velocity of the ball, the angle at which it was thrown, and air resistance. However, if we assume that the ball was thrown straight down with no air resistance, the speed when it hits the ground would be equal to the initial velocity.

2. Does the mass of the ball affect its speed when it hits the ground?

According to Newton's Second Law of Motion, the mass of an object does not affect its speed when falling due to gravity. Therefore, the mass of the ball would not affect its speed when it hits the ground.

3. How can we calculate the speed of the ball when it hits the ground?

To calculate the speed of the ball when it hits the ground, we can use the equation v = √(u^2 + 2gh), where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity (9.8 m/s^2), and h is the height from which the ball is dropped.

4. Does the height at which the ball is dropped affect its speed when it hits the ground?

Yes, the height at which the ball is dropped does affect its speed when it hits the ground. The higher the ball is dropped from, the greater its speed will be when it hits the ground due to the increased potential energy it gains from being at a higher height.

5. Can the speed of the ball when it hits the ground be greater than its initial speed?

No, the speed of the ball when it hits the ground cannot be greater than its initial speed. This is because of the conservation of energy, which states that energy cannot be created or destroyed, only transferred or transformed. Therefore, the total energy of the ball will remain the same throughout its motion, and the speed when it hits the ground will be equal to or less than its initial speed.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
305
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
658
Replies
4
Views
1K
Back
Top