Question

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.
(a) What was the takeoff speed for such a leap?
(b) What horizontal distance did the froghopper cover for this world-record leap?

Answer 1

Answer:

0.528m

Explanation:

a)58.7 cm = 0.587 m

Let g = 9.8m/s2. When the frog jumps from ground to the highest point its kinetic energy is converted to potential energy:

$E_p = E_k$

$mgh = mv^2/2$

where m is the frog mass and h is the vertical distance traveled, v is the frog velocity at take-off

$v^2 = 2gh = 2*9.8*0.587 = 11.5$

$v = \sqrt{11.5} = 3.4 m/s$

b) Vertical and horizontal components of the velocity are

$v_v = vsin(\alpha) = 3.4sin(58^0) = 2.877 m/s$

$v_h = vcos(\alpha) = 3.4cos(58^0) = 1.8 m/s$

The time it takes for the vertical speed to reach 0 (highest point) under gravitational acceleration g = -9.8m/s2 is

$\Delta t = \Delta v / g = \frac{0 - 2.877}{-9.8} = 0.293s$

This is also the time it takes to travel horizontally, we can multiply this with the horizontal speed to get the horizontal distance it travels

$s_h = v_ht = 1.8*0.293 = 0.528 m$