Physics question........ Help!!!?

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 0.65 s, and passes it again on the way down 1.5 s after it was tossed.

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

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

Answer 1

first you find the total time the ball is in the air. The first segment is until it passes the line, which is .65 seconds. Then it reaches the line again at 1.5 seconds. So 1.5 - .65 gives you .85 second. And then lastly, you'll take the same time to reach the ground from the line as from the ground to the line (a symmetry trick). So the last segment is .65. So total flight time is 2.15 seconds.

one of the physics equation is v = v' - gt

given that the ball will be momentary stop at the top, then it'll take the ball 2.15/2 seconds to reach the top, which is 1.075. v' = initial speed. v = 0 (final velocity, at the top). plug it in v' = -(-9.8)(1.075) = 10.535 m/s and this is your initial speed. The height of the line is the function of the time again. And this time, we'll use the equation x = x' + vt + at^t. x' = 0, t = .65 (as told in the problem), a = -9.8. plug everything in....x = 10.535(.65) - (9.8)(.65)^2 = 6.84775 - 4.1405 = 2.70725m.  

Answer 2

let (u) be intial speed and (h) be height of wire

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you may be knowing that distance travelled under gravity is a function of time (t^2), so the object thrown upwards has (2 times, t1 & t2) to hit (or touch) the same same point (say wire) > one while going up and other while coming down

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h = ut - 0.5 gt^2

gt^2 - 2u t + 2h =0

its a quadretic equ in time. let its roots be t1 & t2

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ax^2 + bx + c =0 has 2 roots given by

sum > x1+x2 = - b/a

product> x1 x2 = c/a

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so> t1+t2 = + [2u/g] = 0.65 + 1.5 = 2.15

t1 t2 = [2h/g] = 0.65*1.5 = 0.975

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for > g=9.8

h = 0.975*9.8/2 = 4.7775 m

u = 2.15*9.8/2 = 10.535 m/s