A particularly athletic frog can push herself off of the ground with enough velo

Question

A particularly athletic frog can push herself off of the ground with enough velocity to jump 1m high. Note that the initial velocity of the frog, once you find it, is a property of the frog: no matter what we do with the frog, the initial velocity (which determines how high she can jump) is the same. 1. With what velocity does the frog leave the ground? 2. How long will the frog be in the air before she lands? Suppose we now put the frog in an elevator, accelerating upward at 2 meters per second squared. There is a tasty spider 85 cm above the floor of the elevator. 3. If the frog jumps as high as she can, will she catch the spider? How far above the elevator floor will the frog make it? 4. How long will the frog be in the air? Hint: Think very carefully about your coordinate system, and all of the consequences of the accelerating elevator. You may need the quadratic formula for this problem. If you are still stuck, draw position vs. time graphs for the frog and the spider.

The same frog is now taken to the planet Twilo, which is quite Earthlike except for its value of g=11.8 m/s^2.

How high can our spacefaring frog jump on Twilo?

Based on a comparison of the two jumping-frog problems, can you make any statements regarding gravity and acceleration?

Show all your work please.

Explanation / Answer

1. height h =1m

g = -9.8 m/s2

v=0

we have

2gh = (V2-Vo2)

Vo2 = V2-2gh

plugging in the values we get

V0 =4.43m/s

2) time spent in air

t = 2V0/g = 0.9s

3) dstance moved by elevator in the given time is

h1 = (1/2)at2 (assumiing elevator starts from rest)

h1 = 0.81m

so the frog can not catch the spider.

frog would be at a height h2 = h-h1 = 1- 0.81 = 0.19 m above elevator

4) relative acceleration

a1 = g+a

t = 2V0/(g+a) = 0.73s

on twilo it would jump to a height

h =Vo2/2g' = 0.78m

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