Initial Speed of Ball Thrown Horizontally from 15.75m Height

In summary, the conversation discusses finding the initial speed of a horizontally thrown ball that hits the ground with a final speed that is 5 times its initial speed. The solution involves using the fact that the initial velocity is purely horizontal, finding the vertical component of the final velocity, and then using the given information to solve for the initial speed. The formula v_x^2+v_y^2 = 25 v_x^2 is used to find the initial speed, which is determined to be 3.6m/s.
  • #1
Eleet
9
0
A ball is thrown horizontally from a height of 15.75 m and hits the ground with a speed that is 5.0 times its initial speed. What was the initial speed?


I thought you might have to find time and I got t=1.79s, but after that I cannot seem find my way to the answer

Thanks to whom ever can help me out,

Eleet
Shaun
 
Physics news on Phys.org
  • #2
The fact that the ball is thrown horizontally tells you that the vertical (y) component of the initial velocity is zero. So the initial velocity is purely x-component. Find the vertical component of the final velocity considering that the ball falls the given distance. (Hint: use a kinematic equation relating speed and distance for accelerated motion.)
 
  • #3
??

I do not know if this right but I found:
Vf=17.58m/s
by using Vf^2=2ad.

Than how do I use this to find initial velocity horizontally. Or is that it?
 
  • #4
the initial velocity will be what ever it is, this same velocity is also part of the final.

so [itex] \Vec {V_f}: [V_x,V_y] = 5\Vec {V_i}: [V_x,0] [/itex]
 
Last edited:
  • #5
ExtravagantDreams said:
the initial velocity will be what ever it is, this same velocity is also part of the final.

so [itex] \Vec {V_f}: [V_x,V_y] = 5\Vec {V_i}: [0,V_x] [/itex]

Be careful! That's 5 times the speed - not velocity! :-)
 
  • #6
right, but the initial speed is a velocity with only one component, namely the x
But I did put them in the wrong order
 
  • #7
More to the point: [tex]v_x^2+v_y^2 = 25 v_x^2[/tex]
 
  • #8
Eleet said:
Vf=17.58m/s
by using Vf^2=2ad.
What you found is the y-component of the final velocity. Good! The final velocity has two components: [itex]v_x[/itex] (which is just the initial speed) and [itex]v_y[/itex] (which you just found).

Than how do I use this to find initial velocity horizontally. Or is that it?
Now apply what was given in the problem statement: that the final speed is 5 times the initial speed. Tide gives you the formula:
Tide said:
More to the point: [tex]v_x^2+v_y^2 = 25 v_x^2[/tex]
Note that the initial speed is [itex]v_x[/itex], which is what you're trying to find.

Do you understand how Tide got his formula?
 
  • #9
undefinedundefinedI got it. :biggrin:

we know height so we need to use this formula:

(5Vo)^2=Vo^2+ g2h
>>> Subtract out the Vo^2
>>>>24Vo^2=2gh
>>>>>Vo^2=2gh /2

then, Vo=3.6m/s :laughing:
 
Last edited:

1. What is the formula for calculating the initial speed of a ball thrown horizontally from a height of 15.75m?

The formula for calculating the initial speed (v0) of a ball thrown horizontally from a height (h) is v0 = √2gh, where g is the acceleration due to gravity (9.8 m/s2).

2. Can the initial speed of a ball thrown horizontally from 15.75m height be negative?

No, the initial speed of the ball cannot be negative since it is a measure of the magnitude of the velocity at the starting point. However, the direction of the velocity can be either positive or negative depending on the direction of the throw.

3. How does the initial speed of a ball thrown horizontally from 15.75m height affect its trajectory?

The initial speed of the ball does not affect its trajectory, as long as it is thrown horizontally. This is because the horizontal component of the initial velocity remains constant throughout the motion, while the vertical component is affected by gravity.

4. Is there a difference in the initial speed of the ball if it is thrown from a height above sea level?

Yes, the initial speed of the ball will be affected by the height above sea level due to the change in gravitational acceleration. The closer the ball is to the Earth's surface, the higher the acceleration due to gravity, resulting in a higher initial speed.

5. How can the initial speed of the ball thrown horizontally from 15.75m height be measured?

The initial speed of the ball can be measured using a stopwatch and a ruler. The time it takes for the ball to reach the ground can be recorded and the distance it travels horizontally can be measured. By plugging these values into the formula v0 = √2gh, the initial speed can be calculated.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
289
  • Introductory Physics Homework Help
Replies
25
Views
470
  • Introductory Physics Homework Help
2
Replies
38
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
271
  • Introductory Physics Homework Help
Replies
34
Views
696
  • Introductory Physics Homework Help
Replies
7
Views
4K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
22
Views
4K
Back
Top