Question

A human expedition lands on an exoplanet. One of the explorers is able to jump a maximum distance of 19.5 m with an initial speed of 2.80 m/s. Find the gravitational acceleration on the surface of the exoplanet. Assume the planet has a negligible atmosphere.

Answer 1

Answer:

Acceleration, $a=-0.20\ m/s^2$

Explanation:

It is given that,

Initial speed of the explorers, u = 2.8 m/s

Finally it lands, v = 0

Maximum distance, x = 19.5 m

We need to find the gravitational acceleration on the surface of the exoplanet. Let a is the acceleration. It can be calculated using third equation of motion as :

$v^2-u^2=2ax$

$a=\dfrac{v^2-u^2}{2x}$

$a=\dfrac{(0-(2.8)^2}{2\times 19.5}$

$a=-0.20\ m/s^2$

So, the gravitational acceleration on the surface of the exoplanet is $0.20\ m/s^2$. Minus sign shows that the acceleration is acting in downward direction. Hence, this is the required solution.

Answer 2

Gravitational acceleration, on the surface,of the exoplanet where human expedition lands  is -0.20 meter per squared second.

What is the equation of motion?

The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.

The third equation of the motion for distance can be given as,

$v^2=u^2+2as$

Here, $u$ is the initial body, $v$  is the final body, $a$ is the acceleration of the body and $s$ is the distance traveled.

It is given in the problem,

The maximum distance jumped by one of the explorers is 19.5 meters.

The initial speed is 2.80 m/s.

The gravitational acceleration on the surface of the exoplanet has to be find out.

At this point the velocity of the explorers is zero. Thus by the third equation of motion,

$(0)^2=(2.8)^2+2a(19.5)\\a=-0.20\rm m/s^2$

Here, the negative sign indicates the direction of the explorers is downward.

Thus, the gravitational acceleration on the surface of the exoplanet is -0.20 meter per squared second.

Learn more about the equation of motion here;

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