Kinematics free fall problems part 3 help

AI Thread Summary
The discussion focuses on solving kinematics problems related to a ball thrown upward that passes an overhead power line. Initially, the user calculated the ball's initial speed as 10.29 m/s and the height of the power line as 5.40 m, but both answers were incorrect. After receiving feedback, the user corrected the initial speed to 10.04 m/s but struggled with the height calculation. It was clarified that the power line must be below the ball's maximum height since it passes the line twice. The user then recalculated the height using the correct initial speed and time, arriving at approximately 4.994 m.
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Homework Statement


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 0.85 s, and passes it again on the way down 1.2 s after it was tossed.

(a) What is the height of the power line?

(b) What is the initial speed of the ball?

Homework Equations


v = v0 + at
x=x0+v0*t + 1/2*a*t^2
x = x0 +v*t - 1/2*a*t^2
v^2 = v0^2 +2*a*(x-x0)


The Attempt at a Solution



I drew a picture and decided to solve for intial speed first. I used the equation v = v0 + a* t . V=0, a = -9.8m/s^2 and i took time at v0 to be 1.25 + 0.85/2 = 1.05 s
V0 = 10.29m/s

I then solved part a by using V^2 = V0^2 + 2 *a*(x-x0)
0=(10.29)^2 +2(9.8)(x-0)
i solved this and ended up with 5.40m

I got both of these wrong on my practice test. I would greatly appreciate if someone could explain to me what i did wrong. Thanks so much!
 
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If the time in the problem reads 1.2 s, why does the time in your answer read 1.25 s?
 
Char. Limit said:
If the time in the problem reads 1.2 s, why does the time in your answer read 1.25 s?

oh you are correct! I'm dyslexic! :( I resolved the problem and got 10.04m/s for my initial speed which is correct. However, my distance of 5.148m is still wrong. Any ideas?
 
In your equation you set v equal to 0. This only happens at one point which is when the ball reaches its maximum height. Nowhere is it stated that the power line is at the top of the ball's trajectory. In fact it is clearly stated that the ball passes the power line twice, which can only happen if the power line is below the ball's maximum height.
 
Cyosis said:
In your equation you set v equal to 0. This only happens at one point which is when the ball reaches its maximum height. Nowhere is it stated that the power line is at the top of the ball's trajectory. In fact it is clearly stated that the ball passes the power line twice, which can only happen if the power line is below the ball's maximum height.

you are correct, now that i got my initial velocity of 10.04m/s, g = -9.80m/s^2, x0= 0 and time crossing power line =0.85 i used x = x0 + V0 * t + 1/2*g*t^2
x = 0 + (10.04m/s)(0.85s ) + 1/2 (-9.8m/s^2)(0.85)^2 = 4.994 m How does that look?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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