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. (see nature, vol. 424, july 31, 2003, p. 509.) (a) what was the takeoff speed for such a leap? (b) what horizontal distance did the froghopper cover for this world-record leap?

(a) 4.0 m/s

We can solve this part just by analyzing the vertical motion of the froghopper.

The initial vertical velocity of the froghopper as it jumps from the ground is given by

(1)

where

is the takeoff speed

is the angle of takeoff

The maximum height reached by the froghopper is

h = 58.7 cm = 0.587 m

We know that at the point of maximum height, the vertical velocity is zero:

Since the vertical motion is an accelerated motion with constant (de)celeration , we can use the following SUVAT equation:

Solving for ,

And using eq.(1), we can now find the initial takeoff  speed:

(b) 1.47 m

For this part, we have to analyze the horizontal motion of the froghopper.

The horizontal velocity of the froghopper is

And this horizontal velocity is constant during the entire motion.

We now have to calculate the time the froghopper takes to reach the ground: this is equal to twice the time it takes to reach the maximum height.

The time needed to reach the maximum height can be found through the equation

Solving for t,

So the time the froghopper takes to reach the ground is

And since the horizontal motion is a uniform motion, we can now find the horizontal distance covered:

divide the right side by the left side

0.024÷1.2=0.02 for example

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