Answer
(a) The initial speed is 1.50 m/s.
(b) The height of the cliff is 4.66 meters.
Work Step by Step
(a) $v_y^2 = v_{0y}^2+2gy$
$v_{0y}^2 = v_y^2-2gy = 0-2gy$
$v_{0y} = \sqrt{-2gy} = \sqrt{-(2)(-9.80~m/s^2)(0.0674~m)}$
$v_{0y} = 1.15~m/s$
We can find the initial speed $v_0$.
$\frac{v_{0y}}{v_0} = sin(\theta)$
$v_0 = \frac{v_{0y}}{sin(\theta)} = \frac{1.15~m/s}{sin(50.0^{\circ})}$
$v_0 = 1.50~m/s$
The initial speed is 1.50 m/s.
(b) We can find the time in the air $t$.
$t = \frac{x}{v_x} = \frac{1.06~m}{(1.50~m/s)~cos(50.0^{\circ})}$
$t = 1.10~s$
We can use $t$ to find the height $y$ of the cliff.
$y = v_{0y}~t-\frac{1}{2}gt^2$
$y = (1.15~m/s)(1.10~s)-(4.90~m/s^2)(1.10~s)^2$
$y = -4.66~m$
The grasshopper dropped 4.66 meters from the intial height at the top of the cliff. Therefore, the height of the cliff is 4.66 meters.
