A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.1 m/s. How long does he have to get out of the way if the shot was released at a height of 2.27 m, and he is 1.96 m tall?
We need to use the kinematic equation S=ut+(1/2)at^2 where S=displacement (+=up, in metres) u=initial velocity (m/s) t=time (seconds) a=acceleration (+=up, in m/s^2)
Substitute values S=displacement = 1.96-2.27 = -0.31 m (so that shot does not hit his head) u=11.1 a=-9.81 (acceleration due to gravity)
-0.31=11.1t+(1/2)(-9.81)t^2 Rearrange and solve for t -4.905t^2+11.1t-0.31=0 t=-0.02756 or t=2.291 seconds Reject the negative root to give t=2.29 seconds (to 3 significant figures)
