College Physics (4th Edition)

(a) It takes $1.565~s$ for a golf ball to fall from rest for a distance of 12.0 m (b) In twice that time, the golf ball would fall $48.0~m$
(a) We can find the time it takes for a golf ball to fall from rest for a distance of 12.0 m: $\Delta y = \frac{1}{2}at^2$ $t = \sqrt{\frac{2\Delta y}{a}}$ $t = \sqrt{\frac{(2)(12.0~m)}{9.80~m/s^2}}$ $t = 1.565~s$ It takes $1.565~s$ for a golf ball to fall from rest for a distance of 12.0 m (b) We can find the distance the ball falls in twice that time, that is, in a time of $3.13~s$: $\Delta y = \frac{1}{2}at^2$ $\Delta y = \frac{1}{2}(9.80~m/s^2)(3.13~s)^2$ $\Delta y = 48.0~m$ In twice that time, the golf ball would fall $48.0~m$