## College Physics (4th Edition)

(a) The pole vaulter lands with a speed of $10.8~m/s$ (b) The average force exerted on the pole vaulter during that time interval is $1296~N$
(a) We can find the speed after falling 6.0 meters: $v_f^2 = v_0^2+2ay$ $v_f = \sqrt{v_0^2+2ay}$ $v_f = \sqrt{0+(2)(9.80~m/s^2)(6.0~m)}$ $v_f = 10.8~m/s$ The pole vaulter lands with a speed of $10.8~m/s$ (b) The impulse exerted on the pole vaulter is equal to the change in momentum. We can use the magnitude of the change in momentum to find the average force on the pole vaulter due to the padding: $F~t = \Delta p$ $F~t = m~\Delta v$ $F = \frac{m~\Delta v}{t}$ $F = \frac{(60.0~kg)(10.8~m/s)}{0.50~s}$ $F = 1296~N$ The average force exerted on the pole vaulter during that time interval is $1296~N$.