Tom Throws a Ball: Horizontal & Vertical Velocity

AI Thread Summary
Tom throws a ball from a 45m high cliff with an initial horizontal velocity of 30 m/s. The vertical velocity at impact is calculated to be 30 m/s, while the horizontal velocity remains constant at 30 m/s. The correct time of flight is determined to be 3 seconds, not 4.5 seconds as initially thought. The resultant velocity upon impact is found to be approximately 42.2 m/s at an angle of 45 degrees. This discussion highlights the importance of using the correct equations for projectile motion to arrive at accurate results.
littledude565
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Homework Statement


Tom throws a ball of a cliff 45m nigh, with a velocity of 30 m/s. Take accleration due to gravity as 10m/s. (It took 4.5s to reach the bottom)

What was its horizontal and verticle velocitys when it hit the ground?

i used the first equation (v=u+at) and with substituion i got -42m/s for the verticle velocity. Is this correct, and how would i find the horizontal? thanks for the help
 
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littledude565 said:

Homework Statement


Tom throws a ball of a cliff 45m nigh, with a velocity of 30 m/s. Take accleration due to gravity as 10m/s. (It took 4.5s to reach the bottom)

What was its horizontal and verticle velocitys when it hit the ground?

i used the first equation (v=u+at) and with substituion i got -42m/s for the verticle velocity. Is this correct, and how would i find the horizontal? thanks for the help

Welcome to PF.

If your initial velocity was horizontal, then the time to fall is not given by the velocity relationship. That equation would be useful if you wanted to know how long it took to reach its height i.e when velocity goes to 0.

To find the time you need to use the relationship that relates distance and acceleration and time.

x = 1/2*g*t²
 
the g stands for gravity right? if so would the answer for the verticle be 101.25?
 
and the horizontal 30m/s as its a constant?
 
littledude565 said:
the g stands for gravity right? if so would the answer for the verticle be 101.25?

Based on what?

You'll have to show where that comes from.
 
in your equation x = 1/2*g*t²

So using that i did (assuming the G did stand for gravity). 1/2 * 10m/s*4.5s^2 and i got an answer of 101.25
 
littledude565 said:
in your equation x = 1/2*g*t²

So using that i did (assuming the G did stand for gravity). 1/2 * 10m/s*4.5s^2 and i got an answer of 101.25

You've run out the corral without the saddle there.

The height is given as 45 m. Your time calculation as I said already is not based on the first equation you applied. 4.5 sec is just plain wrong.

So start again with the right equation and find the correct time and then you can figure it out.
 
Ok using the s=1/2*(u+v)*t i did 45=15t, t=45/15 and so i got 3seconds. Is that correct?
 
littledude565 said:
Ok using the s=1/2*(u+v)*t i did 45=15t, t=45/15 and so i got 3seconds. Is that correct?

3 seconds is a much better value to use.
 
  • #10
so what equation do i use now? :S
 
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  • #11
ok so if the verticle velocity increases by 10 every second, would the answer then be 30m/s?
 
  • #12
littledude565 said:
ok so if the verticle velocity increases by 10 every second, would the answer then be 30m/s?

Yes the vertical is 30 m/s at impact. As is the horizontal incidentally.
 
  • #13
ah thank you very much. So would that mean the resultant of these would be 54.08? I got that using phythag sqrt 45^2+30^2 = 54.08 with an angle of 34
 
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  • #14
littledude565 said:
ah thank you very much. So would that mean the resultant of these would be 54.08? I got that using phythag sqrt 45^2+30^2 = 54.08

Why do you think it's 452 again?
 
  • #15
http://img21.imageshack.us/img21/7229/83881376.png
 
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  • #16
ohhh no it should be 45 it should also be 30! So with that now would the resultant be 42.2, with an angle of 45?
 
  • #17
http://img24.imageshack.us/img24/9337/71116538.png
 
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  • #18
Yes. That would be better now.
 
  • #19
Ah yay, thanks for all your help
 
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