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ichivictus
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Homework Statement
A ball is thrown upwards from a point near the sidewalk next to a tall office building. At some later time an occupant of the building, exactly 5.1 meters above the point from which the ball was released, sees it pass her window with a speed of 2.9 m/s.
Calculate the maximum height reached by the ball.
You may neglect air resistance. Take the acceleration due to gravity to be 9.82 m/s^2
Homework Equations
Kinematic Equations
y = Vot + (1/2)at2
Vf = Vo + at
Vf2 = Vo2 + 2ay
y = t * [ (Vo + Vf) / 2 ]
The Attempt at a Solution
Using Kinematic equation Vf^2=Vo^2+2ay I solve for Vo
(2.9)^2=Vo^2 - 19.64*5.1
Vo = 10.01 m/s
To find max height I set Vf to 0 and solve for y using the same acceleration and the newly found Vo.
0 = 10.01^2 - 19.64y
y=5.1 m
Obviously this doesn't work since 5.1m is where the person at the window sees the ball traveling with velocity. Any help would be great, I tried using the other kinematic equations but it got too messy too quickly.