Answer
(a) $1.87~m/s$
(b) $1.02~m/s$
Work Step by Step
See conversion factors (inside front cover).
$1~ft=0.305~m$
$20.5~ft=(20.5~ft)(\frac{0.305~m}{1~ft})=6.2525~m$
$6.00~ft=(6.00~ft)(\frac{0.305~m}{1~ft})=1.83~m$
First, we need to find the deceleration: make the positive direction the direction from the ball to the hole and the hole the origin.
$v_0=1.57~m/s$, $v=0$, $x_0=-20.5~ft=-6.2525~m$, $x=-6.00~ft=-1.83~m$, $\Delta x=x-x_0=-1.83~m-(-6.2525~m)=4.4225~m$
$v^2=v_0^2+2a\Delta x$
$0^2=(1.57~m/s)^2+2a(4.4225~m)$
$a(-8.845~m)=2.4649~m^2/s^2$
$a=\frac{2.4649~m^2/s^2}{-8.845~m}=-0.2787~m/s^2$
(a) $v=0$, $x_0=-20.5~ft=-6.2525~m$, $x=0$, $\Delta x=x-x_0=0-(-6.2525~m)=6.2525~m$, $a=-0.2787~m/s^2$
$v^2=v_0^2+2a\Delta x$
$0^2=v_0^2+2(-0.2787~m/s^2)(6.2525~m)$
$v_0=±\sqrt {2(0.2787~m/s^2)(6.2525~m)}$
$v_0=1.87~m/s$ (positive direction)
(b) To make the remaing $6.00~ft$ putt:
$v=0$, $a=-0.2787~m/s^2$, $x_0=-6.00~ft=-1.83~m$, $x=0$, $\Delta x=x-x_0=0-(-1.83~m)=1.83~m$
$v^2=v_0^2+2a\Delta x$
$0^2=v_0^2+2(-0.2787~m/s^2)(1.83~m)$
$v_0=±\sqrt {2(0.2787~m/s^2)(1.83~m)}$
$v_0=1.02~m/s$ (positive direction)
