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Question: To find the height of an overhead power line, you throw a ball straight upward. The ball passes the line on the way up after 0.80sec , and passes it again on the way down 1.5sec after it was tossed. What is the height of the power line? What is the initial speed of the ball?

What I know/What I don't get: I am having trouble finding the height of this power line with the given information and the information I conceptually know:

acceleration =-9.8 m/s^2

time of the first pass =0.8 sec

time of the second pass =1.5 sec

height of the power line=?

Initial Velocity = ?

Initial Speed = ?

I suspect I must use the formula $$x= x_0 + v_0 t + \frac{1}{2}at^2$$ but I don't quite know for sure what the two different times should be defined as or what values should I plug in. I asked myself these questions in advance. Should I set the formula equal to each other with the two different times plugged into either the left or right equation? Another thought I have is, should I subtract the two different times and solve from there? Please me know if I am asking the right questions and the content of this post is grammatically correct. If not please edit this post for the sake of the quality of the content.

Thanks to all that help in advance.

Disclaimer: I'm not looking for the answer, I am only looking for help to set me in the right direction to solve this problem on my own to further understand how to deal with a problem like this.

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  • $\begingroup$ Calculate the initial velocity of ball first. If you have trouble with that, how fast do you have a throw a ball so that it lands after 1.5+0.8 seconds? $\endgroup$
    – Kvothe
    Feb 2, 2014 at 22:48
  • $\begingroup$ Must I calculate initial velocity with v = v0 + at or x = x0 + (v0)t + (1/2)at^2? $\endgroup$
    – ToltarTheGreat
    Feb 2, 2014 at 23:10
  • $\begingroup$ Well, isn't the (vertical) speed on the top of the parabola (at half of the total travel time) equal to zero? Which formula gives the answer you need? $\endgroup$
    – Kvothe
    Feb 2, 2014 at 23:12
  • $\begingroup$ Well since the equation x = x0 + (v0)t + (1/2)at^2 has a t^2, the x versus t graph must be a parabola for this equation. Therefore I must solve it with this equation, am I correct? $\endgroup$
    – ToltarTheGreat
    Feb 2, 2014 at 23:15

2 Answers 2

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Forget about the power line at first and try to characterize the flight of the ball. You are right in that the ball will be accellerating downwards at 9.8 m/s2 the entire time it is in the air.

Since you are apparently allowed to ignore air resistance, you should be able to see that the time from throw to top is the same as the time from top to catch, and that it would pass the power line the same time difference before and after it reaches the top. Put another way, it reaches the top half way in time between passing the power line on the way up and down.

From this you can easily find the time at top, which is the time from throw to top. From the accelleration over that time, you can find the initial vertical speed. From the initial speed you can find the height at the top. From the start speed, accelleration, and time to pass the power line on the way up, you know the speed at the time it passed the power line. From that you can calculate the average speed from throw to first power line crossing, from which you can find the height.

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In a vertical throw the time taken to ascend a certain distance is the same as the time taken to the sane distance, from this we can assert that the total time taken of ball reaching max height and landing is $(1.5 + 0.8) s$

Now you can use the equation $$y = v_{0}t-\frac{g t^2}{2}$$ and put $y=0$ for total journey time say $T$

This would give you the initial velocity as $$v_{0} = \frac{gT}{2}$$

Now you can again use the initial velocity and time taken to reach the wire to get the height of the wire !

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  • $\begingroup$ Just to let you know, you spelled same wrong when you said "taken to the sane" in the first sentence. Just pointing that out. $\endgroup$
    – ToltarTheGreat
    Feb 2, 2014 at 23:47

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