Answer
(a) The takeoff speed is 4.00 m/s
(b) The froghopper covered a distance of 1.47 meters.
Work Step by Step
(a) $v_y^2 = v_{0y}^2+2ay$
$v_{0y} = \sqrt{-2ay} = \sqrt{-(2)(-9.80~m/s^2)(0.587~m)}$
$v_{0y} = 3.39~m/s$
We can find the takeoff speed $v_0$.
$\frac{v_{0y}}{v_0} = sin(\theta)$
$v_0 = \frac{v_{0y}}{sin(\theta)} = \frac{3.39~m/s}{sin(58.0^{\circ})}$
$v_0 = 4.00~m/s$
The takeoff speed is 4.00 m/s
(b) $range = \frac{v_0^2~sin(2\theta)}{g}$
$range = \frac{(4.00~m/s)^2~sin(2\times 58.0^{\circ})}{9.80~m/s^2}$
$range = 1.47~m$
The froghopper covered a distance of 1.47 meters.
