Find the acceleration due to gravity?

Question

You land on an unknown planet somewhere in the universe that clearly has weaker gravity than Earth. To measure g on this planet you do the following experiment: A ball is thrown upward from the ground. It passes a windowsill 15.0 m above ground and is seen to pass by the same windowsill 2.00 s after it went by on its way up. It reaches the ground again 5.00 s after it was thrown. Calculate the magnitude of g (the acceleration due to gravity) at the surface of this planet.

Help?

Answer 1

y=yo+vot-0.5gt^2

for the entire trip y=yo=0

0=vot-0.5gt^2

0=t(vo-0.5gt), we know t=5 sec, and t≠0 so the only solution is

vo-0.5gt=0

vo-0.5g(5)=0

vo-2.5g=0

g=vo/2.5

So we need vo...

For the windowsill

y=yo+vot-0.5gt^2 on the way up and

y=yo+vo(t+2)-0.5g(t+2)^2 on the way down

yo=0 and y=15.0 m

15=vot-0.5gt^2

15=vo(t+2)-0.5g(t+2)^2

Eliminate t from these equations... you'll have vo in terms of g, then take this to our original equation.

15=vot-0.5gt^2

15=vot+2vo-0.5g(t^2+4t+4)

15=vot-0.5gt^2

15=vot+2vo-0.5gt^2-2gt-2g

Subtracting eq 1 from eq 2

0=2vo-2gt-2g

0=vo-gt-g

gt=vo-g

t=vo/g-1

Putting this into 15=vot-0.5gt^2

15=vo(vo/g-1)-0.5g(vo/g-1)^2

15=vo^2/g-vo-0.5g(vo^2/g^2-2vo/g+1)

15=vo^2/g-vo-0.5vo^2/g+vo-0.5g

15=0.5vo^2/g-0.5g

15+0.5g=0.5vo^2/g

30+g=vo^2/g

vo^2=30g+g^2

vo=+√(30g+g^2)

Finally putting it together

g=vo/2.5

g=√(30g+g^2)/2.5

2.5g=√(30g+g^2)

6.25g^2=30g+g^2

5.25g^2-30g=0

g(5.25g-30)=0

since g≠0, 5.25g-30=0

5.25g=30

g=30/5.25

g=5.71 m/s^2

Answer 2

- Sketch trajectory vs time

- y=ax^2+bx+c curve passes through (0,0) therefore c=0

- curve passes points (1.5,15), (3.5, 15), (5, 0), (0, 0)

sub point into equation to solve a and b.

15= a(a.5)^2+b(1.5)  and 15= a(3.5)^2+b(3.5)

solve simultaneous equation

a=-20/7     and b=100/7

put values into motion equation

s=ut+0.5at^2  (remember to multiply a x 2)

  =(100/7)t+0.5(40/7)t^2

from this you can see that the acceleration due to gravity

is -40/7 or........ -5.71 ms^-2 or......... 5.71 ms^-2 downwards

Answer 3

This is a little above me but from the way you asked the question I think the answer is in"m per sec/per sec"  So Brian I think your figure for Vo is initial velocity. but hey is been 25 years since I failed Calculus  so don't listen to me

resend I opened my mouth to soon lol