Physics, 23.09.2019 01:10 yungkxng57

Acelebrated mark twain story has motivated contestants in the calaveras county jumping frog jubilee, where frogs jump as long as 2.2 m have been recorded. if a frog jumps 2.2 m and the launch angle is 45°, find (a) the frog’s launch speed and (b) the time the frog spends in the air. ignore air assistance.

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Answer from: amulets8409

Answers:

a) 6.566 m/s

b) 0.947 s

Explanation:

The frog's jump can be modeled as projectile motion and the main equations that will be useful in this situation are:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (1)

$V^{2}={V_{o}}^{2} - 2gy$ (2)

Where:

is the height of the frog at a given time

$y_{o}=0$ is the initial height of the frog (assuming it has jumped from ground)

$V_{o}$ is the launch speed (initial speed) of the frog

$\theta=45\°$ is the angle at which the frog jumped

is the time the frog spends on air during the whole parabolic movement

$g=9.8m/s^{2}$ is the acceleration due gravity

is the final speed of the frog at a given time

Knowing this, let's begin with the answers:

(a) The frog’s launch speed

In this case we will use equation (2) assuming the final velocity of the frog when it hits the ground is zero ():

$0={V_{o}}^{2} - 2gy$ (3)

Then we will isolate $V_{o}$ :

$V_{o}=\sqrt{2gy}$ (4)

$V_{o}=\sqrt{2(9.8m/s^{2})(2.2 m)}$ (5)

$V_{o}=6.566 m/s$ (6) This is the frog's launch speed

(b) The time the frog spends in the air

Since we already have the value of $V_{o}$ we can find the time the frog spends in the air with equation (1), which is fullfilled when y=0 :

$0=V_{o}sin\theta t-\frac{gt^{2}}{2}$ (7)

Isolating :

$t=\frac{2V_{o}sin\theta}{g}$ (8)

$t=\frac{2(6.566 m/s)sin(45\°)}{9.8m/s^{2}}$ (9)

Finally:

t=0.947 s (10) This is the time the frog spends in the air

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

Answer from: Quest

B? i think so but hope that's right and answers your question

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Acelebrated mark twain story has motivated contestants in the calaveras county jumping frog jubilee,...