physics 12
posted by priya on .
im stuck i have no clue where to start, please help me i just know the givens and some claculations.
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!!

The 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