Physics, 23.03.2020 23:58 gloriuos3841

In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the greatest horizontal distance. Suppose that on earth she is in the air for time t , reaches a maximum height h, and achieves a horizontal distance D.

A) If she jumped in exactly the same way during a competition on Mars, where g(Mars) is 0.379 of its earth value, find her time in the air.
Express your answer in terms of t.
B)If she jumped in exactly the same way during a competition on Mars, where g(Mars) is 0.379 of its earth value, find her maximum height.
Express your answer in terms of h.
C) If she jumped in exactly the same way during a competition on Mars, where g(Mars) is 0.379 of its earth value, find her horizontal distance.
Express your answer in terms of D .

Answers: 3

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Answers

Answer from: jadalawler3

A) 2.64t

B) 2.64h

C) 2.64D

Explanation:

The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The time of flight of a projectile can be found from the equations of motion, and it is found to be

$t=\frac{2u sin \theta}{g}$

where

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the time of flight is .

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:

$t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t$

The maximum height reached by a projectile can be also found using the equations of motion, and it is given by

$h=\frac{u^2 sin^2\theta}{2g}$

where

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the maximum height is .

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:

$h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h$

The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by

$D=\frac{u^2 sin(2\theta)}{g}$

where:

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the horizontal distance covered is D.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:

$D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D$

Answer from: Quest

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

time (seconds)

answer: 0 meters/second^2

Answer from: Quest

The next element is345640

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