Answer
(a) $15\frac{m}{s}$
(b) $43m$
Work Step by Step
(a) We know that
$\frac{1}{2}mv_i^2+mgh=\frac{1}{2}mv_f^2+0$
This can be rearranged as:
$v_i=\sqrt{v_f^2-2gh}$
We plug in the known values to obtain:
$v_i=\sqrt{(29)^2-2(9.81)(32)}$
$v_i=15\frac{m}{s}$
(b) According to the law of conservation of energy
$K.E_{max}+U_{max}=K.E_f+U_f$
$\implies 0+mgy_{max}=\frac{1}{2}mv_f^2+0$
This simplifies to:
$y_{max}=\frac{v_f^2}{2g}$
We plug in the known values to obtain:
$y_{max}=\frac{(29)^2}{2(9.81)}$
$y_{max}=43m$
