How Do You Calculate the Acceleration Due to Gravity on an Unknown Planet?

[Melchior25]

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.

I drew the picture and started to sole the problem this way but I keep getting the wrong answers. Any input would be greatly appreciated.

to me the initial velocity after 2 seconds should be
v=distance/time which is 15m/2sec so the initial velocity should be 7.5m/s at that point. So from here I used

xf=1/2at^2+vit+xi

15=1/2a(2)^2+(7.5)(2)+0

a=-7.5

I get a gravitational acceleration of -7.5m/s^2. I tried 7 a well thinking the magnitude would be positive. But both answers are wrong.

 

[Shooting Star]

to me the initial velocity after 2 seconds should be
v=distance/time which is 15m/2sec so the initial velocity should be 7.5m/s at that point.

The particle is undergoing accn; it's not traveling at a uniform velo. What you have used is valid if speed is const. Correct it.

EDIT:

After another read, it's never even said anywhere that it took 2 secs to travel 15 m. It crossed the window sill at t=t' traveling up and t=t'+5 going down.

Use the eqns for motion under unifrom accn.

 

[Moetasim]

Hello,

Thank you for reaching out for help with your acceleration problem. I am happy to assist you in finding the correct answer.

First, let's review the equation you are using: xf = 1/2at^2 + vit + xi. This is the equation for displacement, not acceleration. To calculate acceleration, we use the equation a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is time.

Using the information given in the problem, we can calculate the final velocity:
vf = vi + at
vf = 7.5m/s + a(5s)

Next, we can calculate the initial velocity using the same equation:
vi = vf - at
vi = (7.5m/s + a(5s)) - a(2s)
vi = 7.5m/s + 5as - 2as
vi = 7.5m/s + 3as

Now, we can use the equation for displacement to find the acceleration:
xf = 1/2at^2 + vit + xi
15m = 1/2a(5s)^2 + (7.5m/s + 3as)(2s) + 0
15m = 25a + 15as + 15m
25a + 15as = 0
a(25 + 15s) = 0
a = 0 or s = -25/15

Since the problem states that the ball is thrown upward and then falls back to the ground, we know that the acceleration cannot be 0. Therefore, we use the value for s to find the acceleration:
a = -25/15
a = -1.67m/s^2

The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, which is upward. This means that the magnitude of the acceleration is 1.67m/s^2.

I hope this helps you solve your problem. If you have any further questions, please let me know.

Best,