Asked by Anonymous
A baseball is thrown horizontally off a cliff with a speed of 24 m/s. What is the horizontal distance, to the nearest tenth of a meter, from the face of the cliff after 2.8 seconds?
To the nearest tenth of a meter, how far has it fallen in that time?
So is 24 m/s Vx or Vox? Do I need to find a theta? How do I do this?? Sorry I know this is probably easy to you but I'm new to physics and I do not understand any of this chapter
To the nearest tenth of a meter, how far has it fallen in that time?
So is 24 m/s Vx or Vox? Do I need to find a theta? How do I do this?? Sorry I know this is probably easy to you but I'm new to physics and I do not understand any of this chapter
Answers
Answered by
Damon
It is thrown horizontally. There is no horizontal force. Therefore it keeps going at 24 m/s in the horizontal direction. Thus
Vx = Vox = 24 m/s
horizontal distance after 2.8 = 2.8 * 24
The vertical problem is entirely separate.
It has no initial speed down.
therefore
distance down = 0 + 0 t + (1/2)(9.8)(2.8)^2
assuming it has not hit the ground by then of course.
Vx = Vox = 24 m/s
horizontal distance after 2.8 = 2.8 * 24
The vertical problem is entirely separate.
It has no initial speed down.
therefore
distance down = 0 + 0 t + (1/2)(9.8)(2.8)^2
assuming it has not hit the ground by then of course.
Answered by
Anonymous
Yeah...thats correct... But don't worry...every1 goes throug the same problem:-) I too went through that situation:-) be cool:)
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