Straight line motion problem:

In summary: KCIn summary, to solve for the initial velocity of the second ball, you must add 1 second to 2.02s (t-down for ball #1) to get 14.6m.
  • #1
xarmenx
2
0
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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  • #2
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
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
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
 

What is a straight line motion problem?

A straight line motion problem is a type of physics problem that involves the motion of an object along a straight line. It typically involves determining the position, velocity, and/or acceleration of an object over a given period of time.

What are the key equations used to solve straight line motion problems?

The key equations used to solve straight line motion problems are the equations for displacement (d = vt), velocity (v = d/t), and acceleration (a = (vf - vi)/t), where d is displacement, v is velocity, a is acceleration, t is time, vf is final velocity, and vi is initial velocity.

How do I know if a straight line motion problem involves constant acceleration?

If the acceleration of the object remains constant throughout the motion, then it is considered a straight line motion problem with constant acceleration. This can be determined by looking at the given information and seeing if the acceleration is changing or staying the same.

What are the units used for displacement, velocity, and acceleration in straight line motion problems?

The units used for displacement are typically meters (m), for velocity it is meters per second (m/s), and for acceleration it is meters per second squared (m/s^2).

How can I use a graph to solve a straight line motion problem?

A graph can be used to solve a straight line motion problem by plotting the data given and then using the slope of the line to determine the velocity and acceleration of the object. The area under the graph can also be used to determine the displacement of the object.

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