Answer
a) $8.6m$
b) $47.0m$
Work Step by Step
a) $(v_f)^2=(v_i)^2+2a\Delta x$
$\Delta y=\frac{0-(25\frac{m}{s}\sin(28^o))^2}{2(-9.8\frac{m}{s^2})}=2.34m$
$t=\frac{2\Delta y}{v_i+v_f}=\frac{2(2.34m)}{25\frac{m}{s}\sin(28^o)}=0.691s$
$\Delta x=(25\frac{m}{s})\cos(28^o)(0.691s)(2)=30.5m$
$m_Sv_S+m_Pv_P=m_Sv'_S+m_Pv'_P$
$(0.25kg)(0\frac{m}{s})+(0.015kg)(230\frac{m}{s})=(0.265kg)v$
$v=13.01\frac{m}{s}$
$\Delta y=\frac{0-(13.01\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=8.64m$
b) $\Delta y=v_it+\frac{at^2}{2}$
$-2.34m=(13.01\frac{m}{s})t+\frac{(9.8\frac{m}{s^2})}{2}t^2$
$t=2.82s$
$25\frac{m}{s}\cos(28^o)(2.82s)=62.2m-15.25m=47.0m$
