A man jumps to a vertical height of 2.5 m. How long was it in the air before returning to Earth?

Question

Known wrong answers
.5
1
1.3
1.38
1.4
.7
3.5

Known wrong answers
.5
1
1.3
1.38
1.4
.7
3.5

I know this because these are the answers i have attempted already...it is a webassign so it check the answer for you!

I take impeccable notes and stay perfectly attentive in class. Yet, I have a 39/40 on this webassign and for some reason it this question has stumped me...

Thanks richuncle the rounding thing killed me all i needed to do was show a little more decimal places!

Answer 1

jumping up with equation vf^2=vi^2+2ax velocity at highest point is 7 m/s
using vf=vi+at t is .7 to go up going down is same so 2(.7)

1.4 seconds
the person above me obviously does not know earth's gravity
try without rounding1.4278s

Answer 2

This is a question of an object starting at an unknown velocity and decellerating at a known rate (gravity, 9.8 m/s^2) to a stop, then re-accellerating until it reaches it's starting position.

Since decelleration is just negative accelleration, we can use the formula for distance travelled under accelleration:

d = 1/2 a * t^2

We know distance d (2.5 meters, a given) and accelleration a (9.8 m/s^2, the value of gravity), so:

2.5m = 1/2 * 9.8m/s^2 * t^2
2.5m = 4.9m/s^2 * t^2

Solving for t, you get:

2.5m = 4.9m/s^2 * t^2
2.5m / 4.9m/s^2 = t^2
2.5m / 4.9m/s^2 = t^2
.510s^2 = t^2

Take the square root of both sides to find t:

sqrt(.510s^2) = sqrt(t^2)
.714s = t

That's the time for him to go upwards, decellerating all the way, until he reaches a stop. Then he accellerates back down to the starting position on the same amount of time, so the total time in the air is:

.714s * 2 = 1.43s

Now technically this number I just came up with is not right, because accelleration due to gravity is really 9.81 m/sec^2 and I did not follow proper rules for significant figures. Once you re-do the math with the right numbers, you should have the right answer.

Answer 3

But on the important side, stay awake in class, take notes and work out the arithmetic yourself.

Answer 4

to rich uncle: YOU OVERLOOKED THE BASIC THREE VARIABLE DISTANCE FORMULA!
From 2.5 meters the rate of descent would be 9.8m/sec, the speed of gravity. It is richuncle who lacks the skill in figuring out what i have to say next.
"d=vt" - distance formula.
change it around for time, and you get "t=d/v". what is d/v? its distance divided by velocity.
the velocity is 9.8/sec, and the distance to travel is 2.5m.
So, 2.5m divided by 9.8m/sec=.25510204....
The meters cancel out and you are left with time. .25510204 is time.

The answer to your question is that he was in the air for .2551 secs.

Answer 5

To jump that high he must have been on the moon. He wouldn't return to earth because he would burn up on re-entry.

If he did land back on the moon it would be 3.125 seconds later.

Answer 6

5 metres