physics
posted by Anonymous .
A shotputter throws the shot with an initial speed of 15.5m/s at a 34 degree angle to the horizontal. calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.20m above the ground.

Calculate the time in the air, T, by solving
2.20 + 15.5*sin34*T 4.9 T^2 = 0
2.20 + 8.667 T 4.9 T^2 = 0
(Use the quadratic equation and take the positive root)
Once you have T, multiply it by the (constant) horizontal velocity component, 12.85 m/s, for the distance travelled. 
paul konerko hits a baseball in such a way that it leaves his bat traveling at 43.4m/s at an angle os 28 degrees above the horizontal when hit the baseball is 1.2 m above the ground 127 meter away is a 3.5 m tall fence iis this is home run