Answer
(a) $v_{0x} = 20.0~m/s$
$v_{0y} = 34.6~m/s$
(b) $t = 3.53~s$
(c) $y_{max} = 61.1~m$
(d) $range = 141~m$
(e) $a_x = 0$
$a_y = 9.80~m/s^2$ (directed straight down)
$v_x = 20.0~m/s$
$v_y = 0$
Work Step by Step
(a) $v_{0x} = v_0~cos(\theta) = (40.0~m/s)~cos(60.0^{\circ})$
$v_{0x} = 20.0~m/s$
$v_{0y} = v_0~sin(\theta) = (40.0~m/s)~sin(60.0^{\circ})$
$v_{0y} = 34.6~m/s$
(b) $t = \frac{v_y-v_{0y}}{g} = \frac{0-34.6~m/s}{-9.80~m/s^2}$
$t = 3.53~s$
(c) $y_{max} = \frac{v_y^2-v_{0y}^2}{2g}= \frac{0-(34.6~m/s)^2}{(2)(-9.80~m/s^2)}$
$y_{max} = 61.1~m$
(d) $range = \frac{v_0^2~sin(2\theta)}{g} = \frac{(40.0~m/s)^2~sin(2\times 60.0^{\circ})}{9.80~m/s^2}$
$range = 141~m$
(e) At the highest point:
$a_x = 0$
$a_y = 9.80~m/s^2$ directed straight down
$v_x = 20.0~m/s$
$v_y = 0$
