- #1
MattPalmer
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A ball is tossed from an upper story window of a building. It has an initial velocity of 8.00 m/s at an angle of 20 degrees below the horizontal. How long does it take the ball to reach a point 10.0m below the level of launching.
I tried using the equation
yf=vi(sin(theta))-.5gt^2
Which I simplified to
(yf-vi(sin(theta)))/.5g = t^2
Then
(10-8(sin(20)))/(4.91) =t^2
Finally t=1.23, however my book says I should be getting 1.18s, what am I doing incorrectly?
Thanks in advance
I tried using the equation
yf=vi(sin(theta))-.5gt^2
Which I simplified to
(yf-vi(sin(theta)))/.5g = t^2
Then
(10-8(sin(20)))/(4.91) =t^2
Finally t=1.23, however my book says I should be getting 1.18s, what am I doing incorrectly?
Thanks in advance