
#1
Jan1808, 06:02 PM

P: 79

1. The problem statement, all variables and given/known data
Nicole throws a ball straight up. Chad watches the ball from a window 5.0m above the point where Nicole released it. The ball passes Chad on the way up, and it has a speed of 10 m/s as it passes him on the way back down. How fast did Nicole throw the ball? 2. Relevant equations Vf=Vi + at y=Vi(t)+.5a(t)(^2) Vf^2=Vi^2+2(a)(y) 3. The attempt at a solution I've tried starting with the final V =0 and I've tried to solve the downward motion of the ball first, to attempt to get maybe the y value, or height of the ball at it's max. Please, where should I start? What am I missing? 



#2
Jan1808, 06:13 PM

Sci Advisor
HW Helper
P: 8,961

A hint, since no energy is lost  the speed going back down at a point is the same as the speed going up at that same height.
You just need v^2 = u^2 + 2as be careful of the signs! 



#3
Jan1808, 07:14 PM

P: 79

Thanks mgb_phys! I finally got the correct answer of initial velocity=14 m/s.
for v^2 = u^2 + 2as; I let the 10 m/s of the downward velocity, u, = + 10 m/s at the same height of 5m on the upward velocity. I let v = unknown variable and the acceleration was equal to 9.8 m/s^2, and s=5m for the height. Thanks for the explanation! 


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