Initial Speed of Ball Thrown Horizontally from 15.75m Height

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Homework Help Overview

The problem involves a ball thrown horizontally from a height of 15.75 m, with the final speed being 5 times the initial speed. Participants are exploring the relationship between the vertical and horizontal components of motion to determine the initial speed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial conditions of the problem, noting that the vertical component of the initial velocity is zero. They explore kinematic equations to relate the vertical motion to the final speed.

Discussion Status

Some participants have provided insights into the relationship between the components of velocity and have derived expressions involving the final and initial speeds. There is an ongoing exploration of how to connect the vertical and horizontal components to find the initial speed.

Contextual Notes

Participants are working under the assumption that the final speed is 5 times the initial speed, and they are using kinematic equations to analyze the motion. There is some confusion regarding the application of these equations and the interpretation of velocity versus speed.

Eleet
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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
 
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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.)
 
??

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?
 
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:
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! :-)
 
right, but the initial speed is a velocity with only one component, namely the x
But I did put them in the wrong order
 
More to the point: [tex]v_x^2+v_y^2 = 25 v_x^2[/tex]
 
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?
 
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:

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