Initial Speed of Ball Thrown Horizontally from 15.75m Height

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SUMMARY

A ball thrown horizontally from a height of 15.75 meters hits the ground with a final speed that is five times its initial speed. The vertical component of the final velocity was calculated to be 17.58 m/s using the kinematic equation Vf² = 2ad. By applying the relationship between the final speed and the initial speed, the initial horizontal speed was determined to be 3.6 m/s. This calculation utilized the formula (5Vo)² = Vo² + 2gh, leading to the conclusion that the initial speed is 3.6 m/s.

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

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

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: v_x^2+v_y^2 = 25 v_x^2
 
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: v_x (which is just the initial speed) and v_y (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: v_x^2+v_y^2 = 25 v_x^2
Note that the initial speed is v_x, 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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