Julia jumps straight upward on mars, where the acceleration due to gravity is 3.7\,\dfrac{\text{m}}{\text{s}^2}3.7 s 2 m 3, po

Question

3 years ago

7

Julia jumps straight upward on mars, where the acceleration due to gravity is 3.7\,\dfrac{\text{m}}{\text{s}^2}3.7 s 2 m 3, po

int, 7, space, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction downward. After 3\,\text{s}3s3, space, s, julia is falling downward with a velocity of 3.1\,\dfrac{\text{m}}{\text{s}}3.1 s m 3, point, 1, space, start fraction, m, divided by, s, end fraction.

Physics

1 answer:

liberstina [14] · Answer 1 · 2022-08-21T17:05:10+03:00

We know the equation of motion v = u+ at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

In this case Final velocity = -3.1 m/s, negative sign indicates it is pointing downward.

acceleration = - 3.7m/ $s^2$ . Negative means acceleration is towards center of planet Mars.

Time taken = 3 seconds

$v = u + at\\ \\-3.1 = u-3.7*3\\ \\u = 8 m/s$

So jumping velocity of Julia = 8 m/s

Julia jumps straight upward on mars, where the acceleration due to gravity is 3.7\,\dfrac{\text{m}}{\text{s}^2}3.7 s 2 m ​ 3, po

Julia jumps straight upward on mars, where the acceleration due to gravity is 3.7\,\dfrac{\text{m}}{\text{s}^2}3.7 s 2 m 3, po