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Initial speed and time

  • Thread starter eophysics
  • Start date
  • #1
2
0

Homework Statement


OK, here's the problem: A ball (first ball) is dropped from rest from a height 20.0m above the ground. Another ball is thrown (second ball) vertically upward from the ground at the instant the first ball is released. Determine the intial speed of the second ball if the two balls are to meet at a height 10.0m above the ground.


Homework Equations


The equations I am using are time = square root(2y/-g), and initial velocity = x/t.


The Attempt at a Solution


For time I get square root((2(-10m))/9.8m/s^2) = 1.43 seconds.
For initial velocity of the second ball I get 10m/1.43s = 6.99m/s.

However, at the end of the problem it says that the answer should be 14m/s but I can't figure out how to get that so am I doing something wrong, or is the answer I am given wrong?
 

Answers and Replies

  • #2
318
0
Welcome to the forum.
The formula v=x/t is only valid if there is no acceleration. Take a look at your textbook again. There should be a formula for velocity when you have acceleration.
 
  • #3
2
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Would that formula be Vy0 = Vy + gt?
 
  • #4
318
0
That depends on how you choose your minuses. The complete form would be
[tex]v_y(t) = v_y(0)+gt[/tex]
 

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