Acceleration due to gravity?

Question

An astronaut on a distant planet wants to determine its acceleration due to gravity. The astronaut throws a rock straight up with a velocity of 16.2 m/s and measures a time of 22.0 s before the rock returns to his hand. What is the acceleration due to gravity on this planet?

I have no clue how to answer this. I tried the equation V=Vo + at, but the answer i get (.736 m/s^2) is not right. How do I do this?

Answer 1

The CORRECT result is t = ~ 1.47 s, exactly TWICE your answer. (Why that is so is explained below.)

You CAN do it with

v = vo - at,

where we use a convention such that ' a ' will be written as positive, even though it is a DOWNWARDS acceleration. (That conforms to us always using ' g ' for the gravitational acceleration on Earth, with the corresponding understanding that we mean the downwards acceleration.)

Having gone UP with + 16.2 m.s, the rock will come back DOWN with v = - 16.2 m/s, so that

- 16.2 = + 16.2 - 22.0 a.

So a = 32.4/22.0 m/s^2= - 1.472727... m/s^2 or - 1.47... m/s^2 to the 3 sig. figs justified by the input data.

I think your problem was probably in putting the final v equal to zero (0), rather than remembering that it would be - 16.2 m/s when the rock returned to his hand. That meant that you were just calculating the time for the rock to rise to its greatest height; and THAT is why you got just HALF of the correct result for the FULL up and down motion.

Live long and prosper.

Answer 2

Drop the ball from varying heights and measure the time it takes to hit the ground. Graph the results. If you have access to a camera that can take a series of shots, take a set in front of a tape measure. If you can use a strobe light that flashes every 1/10 of a second or so, use that. Note that if you drop it from too high, or use too light a ball, air resistance will slow it down and you'll no longer be measuring the acceleration due to gravity. I won't describe it fully. You need to think about that.

Answer 3

Lets simplify this problem further.

Lets assume that the ball was dropped at the highest point and it took 11 seconds to reach the ground with velocity of 16.2

Because using the conservation of energy, the speed of the rock has to be the same at a given point either upwards or downwards.

Also it takes the same time to travel up (11s) and travel down(11s)

Formula
v = u + g*t, in our case u is 0 at its top most point.

16.2 = g*11

g = 16.2/11

g = 1.47 m/s^2

Answer 4

if the time was 22 seconds during the stones upward travel before coming down again. it means that it traveled a short distance.That means the acceleration is quasi constant. So By definition of average acceleration due to gravity we would have velocity divided by the total travel time = 16.2/22. meters per sec^2. The average acceleration is the max acceleration at tthe surface of the Earth divided by two.

Answer 5

upwards.... V(f) = v(i) - gt, but g unknown...
round trip = 22, so falling time = 11 seconds

h = 0.5gt^2 , distance it falls from the highest point , and [ v(f) ]^2 = [ v(i)]^2 + 2 gh , v(f) = -16.2 , the velocity it returns to your hand at, v(i) = 0,the velocity it has as it begins to fall back, ..... replace h with 0.5gt^2 , in the second equation. .... you know t = 11 sec.

you will then need to solve your eqn for g...

I got 1.472727.... as above, .... watch the sig figs in your answer...

chas below should use 11 sec, not 22 sec in his calculation, then he gets the same answer as I do...

Answer 6

How do you get that answer?
V= -16.2
V0= 16.2
because when it returns to him it must have the same SPEED as when it left him. This is IF there is no air friction. This is because energy is conserved.

the answer should be 1.473 m/s^2

Answer 7

I think you have to tackle this in two phases of motion

intial velocity = 16.2
final velocity = 0
time of flight to highest point = ?