#### Answer

(a) The cliff is 26 meters high.
(b) The maximum height is 34 meters.
(c) The impact speed is 20 m/s

#### Work Step by Step

(a) $y = v_{0y}t-\frac{1}{2}gt^2$
$y = (30~m/s)~sin(60^{\circ})(4.0~s)-\frac{1}{2}(9.80~m/s^2)(4.0~s)^2$
$y = 26~m$
The cliff is 26 meters high.
(b) $2ah=v_y^2-v_{0y}^2$
$2(-g)h=v_y^2-v_{0y}^2$
$h=\frac{v_{0y}^2-0}{2g}$
$h=\frac{[v_0~sin(\theta)]^2-0}{2g}$
$h=\frac{[v_0~sin(\theta)]^2}{2g}$
$h=\frac{[(30~m/s)~sin(60^{\circ})]^2}{(2)(9.80~m/s^2)}$
$h = 34~m$
The maximum height is 34 meters.
(c) We can find the horizontal component of velocity.
$v_x = v_0~cos(\theta)$
$v_x = (30~m/s)~cos(60^{\circ})$
$v_x = 15~m/s$
We can find the vertical component of velocity after 4.0 seconds.
$v_y = v_{0y}+at$
$v_y = (30~m/s)~sin(60^{\circ})-(9.80~m/s^2)(4.0~s)$
$v_y = -13.2~m/s$
We can find the impact speed.
$v = \sqrt{(v_x)^2+(v_y)^2}$
$v = \sqrt{(15~m/s)^2+(-13.2~m/s)^2}$
$v = 20~m/s$
The impact speed is 20 m/s