Uncovering the Mass of an Object on Planet X

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The discussion focuses on calculating the weight of a rock on Planet X after it is thrown upward. The initial calculations incorrectly used an acceleration of 8.61 m/s², leading to an erroneous weight of 45.1 N. A more accurate approach involves determining the acceleration using the formula a = 2(s - ut) / t², which yields an acceleration of approximately -17.22 m/s². This value is then used to calculate the weight as F = ma, resulting in a weight of about 90 N. The importance of correctly applying kinematic equations to find acceleration is emphasized for accurate weight determination.
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Homework Statement


When venturing forth on Planet X, you throw a 5.24kg rock upward at 13.0 m/s and find that it returns to the same level 1.51 s later.

What does the rock weigh on Planet X?

Homework Equations



W=m*a

The Attempt at a Solution



To find the acceleration i good the 13.0m/s / 1.51s = 8.61 m/s^2. W=5.24kg * 8.61 m/s^2 = 45.1 N

The online homework keeps telling me that my answer is wrong. What am I doing wrong? Thank you for the help!
 
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Try using this equation:
x = (x-initial) + (V-initial)(t) + (1/2)(a)(t^2)
(remember that in your case the change in x = 0, which means x - (x-initial) = 0)

Someone correct me if I am wrong :0
 
With this formula could be easier: Vf=Vo+at.
Vo is your initial velocity, Vf is the final velocity that for you is the velocity at highest point the stone reaches (which is 0), and the time for you is 1.52/2 because you're just dealing with half of the experiment.
 
You can use: s=ut+\frac{1}{2}at^{2} for the entire 'experiment'.

s=displacement
u=initial (starting) velocity
t=time
a=acceleration

Note: You are trying to find the acceleration so the weight formula can be used later.

Therefore, by rearrangement, a=\frac{2(s-ut)}{t^{2}}

Since the rock is launched at 13ms-1, this is the initial velocity (u)
The time taken for the rock to come back from where it was launched, t=1.51 seconds
This seems to be all the information given that is relevant to this equation. But it is also known that the displacement (s) will be 0 when the rock reaches back to where it left.

Hence, a=\frac{2(0-13(1.51))}{(1.51)^{2}}

a \approx -17.22 ms^{-2} (note: negative acceleration means the acceleration is acting opposite to the rock launch. i.e. downwards towards the planet)

Now since the acceleration has been found, the weight of the rock can be found using:
F=ma

F= force (Newtons) / weight of object
m = mass = 5.24kg
a = acceleration = -17.22 ms-2
Therefore, F=(5.24)(17.22)

F\approx 90kg
 
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