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September 15, 2014

September 15, 2014

Posted by **priya** on Sunday, August 8, 2010 at 12:40pm.

question: kevin stands on he top of a buiulding and he throws an aboject up into the air at a velocity of 16.0m/s at an angle of 25 degrees to the horizontal. if the building is 75 m tall, how far from the base will the object hit the ground.

given:

vi: 16.o m/s

dy: 75m

dx:?

a=9.8m/s^2

calculated given:vix=6.76m/s,viy=14.5m/s

this is all i know, i have know clue what to do, please help me. Thanks!!

- physics 12 -
**drwls**, Sunday, August 8, 2010 at 4:54pmThe vertical component viy should be LESS than the horizontal component, vix. That is because the launch angle is closer to vertical than horizontal.

Use the equation of motion that contains g and viy to compute the time that it takes for the object to hit the ground (y=0).

y = 75 + viy*t - (g/2)*t^2 = 0

Once you have solved for t (using the positive of the two possible solutions), use that and vix to compute the horizontal location.

X = vix*t

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