The froghopper, *Philaenus spumarius*, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizontal, some of the tiny critters have reached a maximum height of 58.7 cm above the level ground.

What was the takeoff speed for such a leap?

What horizontal distance did the froghopper cover for this world-record leap?

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:

- Identify the
and__target variable__for each direction—remember that__known variables__*only*(Δ**3**of the**5**variables*x*or Δ*y*,*v*_{0},*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! __Choose a UAM__—sometimes you'll be able to go directly for the target variable, sometimes another step will be needed in between.**equation**for the target (or intermediate) variable, then**Solve**the equation__substitute known values__and__calculate__the answer.

The four UAM (kinematics) equations are:

$\overline{)\mathbf{}{{\mathit{v}}}_{{\mathit{f}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathbf{}}{\mathbf{+}}{\mathit{a}}{\mathit{t}}\phantom{\rule{0ex}{0ex}}{\mathbf{\u2206}}{\mathit{x}}{\mathbf{=}}{\mathbf{}}\mathbf{\left(}\frac{{\mathit{v}}_{\mathit{f}}\mathbf{+}{\mathit{v}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{\right)}{\mathit{t}}\phantom{\rule{0ex}{0ex}}{\mathbf{\u2206}}{\mathit{x}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathit{t}}{\mathbf{+}}{\frac{1}{2}}{\mathit{a}}{{\mathit{t}}}^{{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{{{\mathit{v}}}_{{\mathit{f}}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{}}{{{\mathit{v}}}_{{\mathbf{0}}}}^{{\mathbf{2}}}{\mathbf{}}{\mathbf{+}}{\mathbf{2}}{\mathit{a}}{\mathbf{\u2206}}{\mathit{x}}}$

In our coordinate system, the **+ y-axis is pointing upwards** and the

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:

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

Projectile Motion: Positive Launch

Projectile Motion: Positive Launch

Projectile Motion: Positive Launch

Projectile Motion: Positive Launch