Question about freely falling objects

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A stone is thrown upwards with an initial speed of 5 m/s from a 30 m high cliff, and the time to reach the ground is calculated using the equation h = Vo t - 0.5 g t^2. The equation simplifies to t^2 - t - 6, which is solved using the quadratic formula, yielding a time of 3 seconds. Some participants suggest considering the upward distance the stone travels before falling, but the consensus confirms that the calculated time is correct. The calculations consistently show that the stone takes 3 seconds to hit the ground.
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Homework Statement


Q: A stone is thrown vertically upwards with initial speed of 5 m/s from edge of a cliff 30 m high. Find the time needed for the stone to reach the ground (in seconds).


Homework Equations


1. (h= Vo t - 0.5 g t^2)
2. Quadratic equation:
3ea647783b5121989cd87ca3bb558916.png



The Attempt at a Solution


First I will use this equation: (h= Vo t - 0.5 g t^2)
-30 = (5) t - 5 t^2
Now I divide by 5 and it will be like this
t^2 - t - 6
Then, I use the quadratic equation
\frac{-(-1)+\sqrt{(-1)^2-4(1)(-6)}}{2(1)}

The answer is 3, is that right?

Another question, I am not sure..
maybe I need to calculate the upward vertical distance it will go first.. then I sum it with the actual height which is 30 m?
 
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RuthlessTB said:
The answer is 3, is that right?
Looks right to me.
 
simple calculation..
considering down as positive...
h=ut+1/2at.t
h=30 m
a= 10 m/s.s
u=-5m/s

30=-5t+5t.t
t=3 s
 
simple calculation..
considering down as positive...
h=ut+1/2at.t
h=30 m
a= 10 m/s.s
u=-5m/s

30=-5t+5t.t
t=3 s
 
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