Sitting in a second-story apartment, a physicist notices a ball moving straight
ID: 585616 • Letter: S
Question
Sitting in a second-story apartment, a physicist notices a ball moving straight upward just outside her window. The ball is visible for 0.25 s as it moves a distance of 1.05 m from the bottom to the top of the window. How long does it take before the ball reappears? Express your answer using two significant figures. What is the greatest height of the ball above the top of the window? Here is a sample answer from a like problem. I can't see why it says divide equation.
u = speed of ball at bottom of window
v = speed of ball at top of window
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v = u - gt = u - 0.25 g
u - v = 0.25 g ---------------- (1)
u - v = 2.45
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v^2 = u^2 - 2 g x = u^2 - 2 g *1.03
u^2 - v^2 = (u - v)(u+v) = 2.06 g -------------- (2)
dividing 2 equs
u + v = 8.24 ------------ (3)
solving (1) & (3)
u = 5.35 m/s
v = 2.9 m/s
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greatest height (H) above window
ME at top of window (origin here) = ME at max height (speed =0)
PE + KE = PE + KE
0 + 0.5 mv^2 = mgH + 0
H = v^2/2g = 2.9*2.9/2*9.8 = 0.43 m
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edit>>>> lets see when velocity becomes zero
g = [0 - v] / t = - 9.8
t = time to reach max height from window top = 2.9/9.8
t = 0.296 s
now it falls freely a distance H = 0.43 m and takes time t1
0.43 = 0 + 0.5 * 9.8 * t1^2
t1 = 0.296
T = time in it reappears = t + t1 = 2*0.296 = 0.592 sec
Explanation / Answer
using the equation S = (u*t)-(0.5*g*t^2)
1.03 = (u*0.25)-(0.5*9.81*0.25^2)
initial velocity is u = 5.34 m/s
greatest height is Hmax = u^2/(2*g) = 5.34^2/(2*9.81) = 1.45 m
from the top of the window h = 1.45-1.03 = 0.42 m
time of ascent is ta = u/g = 5.34/9.81 = 0.54 s
then required time is (0.54-0.25)*2 = 0.58 S