Physics
posted by Jordan .
A shotputter throws the shot (mass = 7.3kg) with an initial speed of 15.0 m/s at a 33.0 degree angle to the horizontal.
Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.00m above the ground.

no air resistance?
consider the horizonal distance:
d= Vihorizontal*time = 15cos33*t
now the vertical:
finalheight=initialheight+ vivertical*t 4.9t^2
or
0=2+15sin33 t  4.9 t^2
solve the second equation for time. Use the quadratic equation. Put that time t into the first equation (distance horizontal). 
sdds

disco
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