Physics problem.......

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An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 15.0 meters if her initial speed in 3.00 m/s. What is the free-fall acceleration on the planet?

Sep 24th, 2015

Thank you for the opportunity to help you with your question!

The key to solve a physics problem is to do this process everytime

1-translate the problem  into variable form

2- choose a formula that involves all the variables

3-Plug in the numbers

Ok here it goes

1-  s =15m      (distance)

     vi = 3 m/s   (initial velocity)   

     a=?            (acceleration)

     vf=0            (final velocity) this one comes from the fact that at the highest point your velocity is 0

2- you need a formula that involves all the variables (s,vi,a,vf)

2as=vf^2-vi^2

3- plug in the numbers

2*a*15=0-3^2

so 

a=0.3


Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 24th, 2015

Thanks

Sep 24th, 2015

Thanks

Sep 24th, 2015

ok I did not see the word horizontal initially that makes a difference

for the maximum horizontal distance you need 45degree shooting so if the velocity is 3m/s she will go 3sin45  vertical and 3cos45 horizontal 

now distance (s) is velocity (v) multiply by time (t)

s=V.t

so s =3cos45.t

the time she has is the time her vertical velocity will buy her

t=(Vf-Vi)/a                      (final velocity -initial velocity)/ acceleration    for only going up which is the same as time  to come down 

t=2*3sin45/a    notice that we multiplied it by 2 to account for the time to go up and back to the ground (all the time she goes up and comes down she moves horizontally)

so s=2*3*3*sin45*cos45/a

or 15=18/2a

so a=0.6

Best

Shirin


Sep 24th, 2015

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