Answer
(a) $v_0 = 12.3~m/s$
(b) The maximum height above the ground is 0.90 meters.
Work Step by Step
(a) $x = \frac{v_0^2~sin(2\theta)}{g}$
$v_0^2 = \frac{gx}{sin(2\theta)}$
$v_0 = \sqrt{\frac{gx}{sin(2\theta)}}$
$v_0 = \sqrt{\frac{(9.80~m/s^2)(10~m)}{sin(2\times 20^{\circ})}}$
$v_0 = 12.3~m/s$
(b) $2a_yh=v_y^2-v_{0y}^2$
$2(-g)h=v_y^2-v_{0y}^2$
$h=\frac{v_{0y}^2-0}{2g}$
$h=\frac{[v_0~sin(\theta)]^2-0}{2g}$
$h=\frac{[v_0~sin(\theta)]^2}{2g}$
$h=\frac{[(12.3~m/s)~sin(20^{\circ})]^2}{(2)(9.80~m/s^2)}$
$h = 0.90~m$
The maximum height above the ground is 0.90 meters.
