For this problem, we're looking for the initial speed and the range of the froghopper's leap given the direction of its launch and the maximum height it has to travel.

Since the takeoff and landing are at the same height, this is a symmetrical launch problem.

For projectile motion problems in general, we'll follow these steps to solve:

1. Identify the target variable and known variables for each direction—remember that only 3 of the 5 variables (Δx or Δy, v₀, v_f, a, and t) are needed for each direction. Also, it always helps to sketch out the problem and label all your known information!
2. Choose a UAM equation—sometimes you'll be able to go directly for the target variable, sometimes another step will be needed in between.
3. Solve the equation for the target (or intermediate) variable, then substitute known values and calculate the answer.

The four UAM (kinematics) equations are:

$$\begin{gathered} \boxed{\mathbf{ }{\mathit{v}}_{\mathit{f}}\mathbf{ }\mathbf{=}\mathbf{ }{\mathit{v}}_{\mathbf{0}}\mathbf{ }\mathbf{+}\mathit{a}\mathit{t} \\ \mathbf{∆}\mathit{x}\mathbf{=}\mathbf{ }\mathbf{\left(}\frac{{\mathit{v}}_{\mathit{f}}\mathbf{+}{\mathit{v}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{\right)}\mathit{t} \\ \mathbf{∆}\mathit{x}\mathbf{=}\mathbf{ }{\mathit{v}}_{\mathbf{0}}\mathit{t}\mathbf{+}\frac{1}{2}\mathit{a}{\mathit{t}}^{\mathbf{2}} \\ \mathbf{ }{{\mathit{v}}_{\mathit{f}}}^{\mathbf{2}}\mathbf{=}\mathbf{ }{{\mathit{v}}_{\mathbf{0}}}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{2}\mathit{a}\mathbf{∆}\mathit{x}} \end{gathered}$$

In our coordinate system, the +y-axis is pointing upwards and the +x-direction is horizontal along the launch direction. That means a_y = −g, and a_x = 0 (because the only acceleration acting on a projectile once it's launched is gravity.)

For the special case of a symmetrical launch, we also have equations for the horizontal distance traveled, also called the range, and maximum height of the projectile:

$\boxed{\mathit{R}\mathbf{=}\frac{{{\mathit{v}}_{\mathbf{0}}}^{\mathbf{2}}\mathbf{ }\mathbf{sin}\mathbf{\left(}\mathbf{2}\mathit{\theta }\mathbf{\right)}}{\mathit{g}}}$  and   $\boxed{{\mathit{H}}_{\mathit{m}\mathit{a}\mathit{x}}\mathbf{=}\frac{{{\mathit{v}}_{\mathbf{0}}}^{\mathbf{2}}\mathbf{ }{\mathbf{sin}}^{\mathbf{2}}\mathit{\theta }}{\mathbf{2}\mathit{g}}}$