Straight line motion problem:

  • Thread starter xarmenx
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I am having a problem solving this:

A ball is thrown straight up from the edge of a 20m tall building (ignore air resistance). One second later a second ball is thrown. What must be the initial velocity of the second ball in order for both of them to hit the ground at the same time.



Knowns:
Ball 1
x=20m
v0=0m/s
a=g (I am using 9.8m/s^2)
t-down=2.02s (x=vt+0.5at^2, 20=0t+0.5(9.8)t^2)

Ball 2
x=20m
v0 is the unknown, t is also unkown
a=g

I don't know how to calculate t-up (ball was thrown up originally) thus I don't know the total time.
The answer in the book is 8.18m/s and t is 1.55s. Any help would be appreciated as to how to obtain those answers.

Using the displacement formula:
x=x0+v0+0.5at^2 I can calculate T-Down for ball #1 which is 2.02s (20=0+0+4.9t^2)

I thought maybe to calculate time down for ball #2 I can subtract 1 second from 2.02s(ball 1 t down) which is 1.02s.
Then substitute that into 20=v0(1.02)+4.9(1.02^2) = 14.6m which is not the correct answer. I assume the error is due to me missing t-up completely.

I also tried solving the equations simoltaneously since they are equal but it seems to make little sense when I do (and I also have 2 unknown on 1 side):

20=v0t+4.9at^2, v0t+4.9at^2-20=0
20=4.9at^2, 4.9at^2-20=0

V0t+4.9at^2-20=4.9at^2-20 (the -20 just cancels and the result is v0t equals 0.
 
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Answers and Replies

  • #2
Hootenanny
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xarmenx said:
I am having a problem solving this:

A ball is thrown straight up from the edge of a 20m tall building (ignore air resistance). One second later a second ball is thrown. What must be the initial velocity of the second ball in order for both of them to hit the ground at the same time.



Knowns:
Ball 1
x=20m
v0=0m/s
a=g (I am using 9.8m/s^2)
t-down=2.02s (x=vt+0.5at^2, 20=0t+0.5(9.8)t^2)

Ball 2
x=20m
v0 is the unknown, t is also unkown
a=g

Welcome to PF,

The question states that "the ball is thrown straight up", however, later you state that the initial velocity of the ball is zero, this cannot be the case if the ball was thrown up, only if it was released from rest and allowed to fall.

~H
 
  • #3
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K guys sorry for the fuss and using up your time but the mistake I was making is like you said the initial ball is thrown up, the second one is dropped. So to get the right answer I had to add that one second not subtract it, which did provide correct results. Thanks again for the help.
 
  • #4
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
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xarmenx said:
K guys sorry for the fuss and using up your time but the mistake I was making is like you said the initial ball is thrown up, the second one is dropped. So to get the right answer I had to add that one second not subtract it, which did provide correct results. Thanks again for the help.

No problem.

~H
 

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