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3 years ago

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Using software, a simulation of a javelin being thrown both on Earth and on the moon is created. In both cases, the velocity and

angle of projection are kept the same. On Earth, the javelin covers a distance y. What distance would it cover on the moon? (The moon’s gravity is about 1/6th that of Earth.)

2 answers:

kodGreya [7K]3 years ago

7 0

$x = (v_{0}) ^{2} * \frac{sin 2 \alpha}{g} + x_{0}$

If g goes down by a factor of 6, the distance x goes up by a factor of 6.

mario62 [17]3 years ago

4 0

Horizontal distance covered by a projectile is X = Vix *T

where Vix is the initial horizontal component of velocity and T is time taken by the projectile

Vix = ViCos theta

In question they said that initial velocity and angle is same on earth and moon

so Vix would remains same

now let's see about time taken T

time taken to reach the highest point

Vfy = Viy +gt

at highest point vertical velocity become zero so Vfy =0

0 = Vi Sin theta + gt

t = Vi Sintheta /g

Total time taken to land will be twice of that

On earth

Te= 2t

Te = 2Sinθ/g

on moon g is one-sixth of g(earth)

Tm = 2Sinθ/(g/6)

Tm = 6(2Sinθ/g)

Tm = 6Te

so total time taken by the projectile on moon will be six times the time taken on earth

From first equation X = Vix*T

we can see that X will also be 6 times on moon than earth

so projectile will cover 6 times distance on moon than on earth

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A roadrunner at rest suddenly spots a rattlesnake slithering directly away at a constant speed of 0.75 m/s. At the moment the sn

Shalnov [3]

Answer:

The roadrunner will take approximately 5.285 seconds to catch up to the rattlesnake.

Explanation:

From the statement we notice that:

1) Rattlesnake moves a constant speed ( $v_{S} = 0.75\,\frac{m}{s}$ ), whereas the roadrunner accelerates uniformly from rest. ( $v_{o, R} = 0\,\frac{m}{s}$ , $a = 1\,\frac{m}{s^{2}}$ )

2) Initial distance between the roadrunner and rattlesnake is 10 meters. ( $x_{o, R} = 0\,m$ , $x_{o,S} = 10\,m$ )

3) The roadrunner catches up to the snake at the end. ( $x_{S} = x_{R}$ )

Now we construct kinematic expression for each animal:

Rattlesnake

$x_{S} = x_{o,S}+v_{S}\cdot t$

Where:

$x_{o, S}$ - Initial position of the rattlesnake, measured in meters.

$x_{S}$ - Final position of the rattlesnake, measured in meters.

$v_{S}$ - Speed of the rattlesnake, measured in meters per second.

$t$ - Time, measured in seconds.

Roadrunner

$x_{R} = x_{o,R} +v_{o,R}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$

Where:

$x_{o, R}$ - Initial position of the roadrunner, measured in meters.

$x_{R}$ - Final position of the roadrunner, measured in meters.

$v_{o,R}$ - Initial speed of the roadrunner, measured in meters per second.

$a$ - Acceleration of the roadrunner, measured in meters per square second.

$t$ - Time, measured in seconds.

By eliminating the final positions of both creatures, we get the resulting quadratic function:

$x_{o,S}+v_{S}\cdot t = x_{o,R}+v_{o,R}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$

$\frac{1}{2}\cdot a \cdot t^{2} + (v_{o,R}-v_{S})\cdot t + (x_{o,R}-x_{o,S}) = 0$

If we know that $a = 1\,\frac{m}{s^{2}}$ , $v_{o, R} = 0\,\frac{m}{s}$ , $v_{S} = 0.75\,\frac{m}{s}$ , $x_{o, R} = 0\,m$ and $x_{o,S} = 10\,m$ , the resulting expression is:

$0.5\cdot t^{2}-0.75\cdot t -10=0$

We can find its root via Quadratic Formula:

$t_{1,2} = \frac{-(-0.75)\pm \sqrt{(-0.75)^{2}-4\cdot (0.5)\cdot (-10)}}{2\cdot (0.5)}$

$t_{1,2} = \frac{3}{4}\pm \frac{\sqrt{329}}{4}$

Roots are $t_{1} \approx 5.285\,s$ and $t_{2}\approx -3.785\,s$ , respectively. Both are valid mathematically, but only the first one is valid physically. Hence, the roadrunner will take approximately 5.285 seconds to catch up to the rattlesnake.

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