The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an

angle of 58.0 degrees 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, 31 July 2003, p. 509.)
(a) What was the take-off speed for such a leap?
(b) What horizontal distance did the froghopper cover for this world record leap?

h = Vi t - 4.9 t^2

.587 = Vi t - .5*9.81 t^2

but t at top is Vi/9.81
where 0 = Vi - g t

.587 = Vi^2/9.81 - .5 * 9.81 (Vi^2/9.81^2)

.587 = .5 * Vi^2/9.81
Vi = 3.39 m/s vertical up

Vi = speed * sin 58
so
speed = 4.00 m/s part a

u = 4.00 cos 58 = 2.12 m/s

time in air = 2 * rise time
rise time = Vi/g
so
time in air = 2 * 3.39/9.81
= .691 seconds in air
range = u * .691
= 2.12 * .691
= 1.47 meters
wow, that is impressive if correct.

4.9 is (1/2)*a, a = gravity which is -9.8

(a) Well, frogs might be known for their leaping abilities, but it seems the froghopper is giving them a run for their money! To find the take-off speed, we'll need to do a little math. Now, considering the angle of 58.0 degrees above the horizontal and a maximum height of 58.7 cm, we can make use of a little projectile motion. So, drumroll, please... the take-off speed for this leap is approximately [calculating]... 71.5 cm/s! That's one speedy insect!

(b) Now, for the horizontal distance covered by our record-breaking froghopper. Hang on tight as we crunch some numbers here. Using the same angle of 58.0 degrees, we can calculate the distance by multiplying the take-off speed by the time of flight. Time for a quick calculation... [calculating]... and we have approximately 94.3 cm! So, the froghopper covered about 94.3 cm horizontally. That's quite the jump for such a little critter!

To answer these questions, we can use the principles of projectile motion. Projectile motion refers to the motion of an object, like the froghopper, that is launched into the air at an angle and experiences only the force of gravity. There are two components of projectile motion: horizontal motion and vertical motion.

(a) To find the take-off speed (initial velocity) of the froghopper, we need to consider the vertical motion component. We are given the maximum height reached by the froghopper, which is 58.7 cm above the level ground.

The maximum height reached by a projectile occurs when its vertical velocity becomes zero. At this point, the gravitational potential energy is maximum and is equal to the kinetic energy it had initially. We can use this principle to find the take-off speed.

The general equation for the maximum height of a projectile is given by:
h_max = (v_i^2 * sin^2θ) / (2g),

where:
- h_max is the maximum height reached,
- v_i is the initial speed (take-off speed),
- θ is the launch angle (58.0 degrees in this case),
- g is the acceleration due to gravity (approximately 9.8 m/s^2).

First, let's convert the maximum height from cm to meters:
h_max = 58.7 cm = 0.587 m.

Plugging in the given values, the equation becomes:
0.587 = (v_i^2 * sin^2(58.0)) / (2 * 9.8).

To solve for v_i, we rearrange the equation and solve for v_i:
v_i = sqrt((0.587 * 2 * 9.8) / sin^2(58.0)).

Calculating this using a calculator, the take-off speed is approximately 2.13 m/s.

(b) To find the horizontal distance covered by the froghopper, we need to consider the horizontal motion component. The horizontal motion is independent of the vertical motion and can be calculated using the formula:

range = (v_i^2 * sin(2θ)) / g,

where range is the horizontal distance covered.

Plugging in the given values, the equation becomes:
range = (2.13^2 * sin(2 * 58.0)) / 9.8.

Calculating this using a calculator, the horizontal distance covered by the froghopper is approximately 0.708 meters or 70.8 cm.

why is a=4.9