Time to Calculate the Piano's Fall

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To calculate the time it takes for a piano to hit the ground after the rope snaps, the initial height is 6.0 m and the initial velocity is 2.9 m/s upwards. Once the rope breaks, the only force acting on the piano is gravity, with acceleration set to -g. The kinematic equation y = y_0 + v_0 t + 1/2 a t^2 can be used to determine the time, leading to the equation 0 = 6 + 2.9t - 4.9t^2. Solving this using the quadratic formula yields a time of approximately 1.441 seconds before the piano reaches the ground.
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A piano is being raised to the third floor of a building using a rope and pulley. The piano is 6.0 m above the ground, moving upwards at 2.9 m s-1, when the rope snaps. Calculate how much time (in s) elapses before the piano hits the ground.

i need to know how to approach the problem
thanks in advance.
 
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Treat the piano like any other falling object. You are given its initial speed and height. (Hint: What's the general kinematic relationship that gives position as a function of time for uniformly accelerated motion?)
 
If we use s = ut + 1/2at^2
for the piano to reach the top
s=6
u=2.9
v=0
a=Don't know
t=?
or
s=0
u=2.9
v=?
a=What would be the a?
t=?
 
Of course you know a! Once the rope snaps, the piano is a freely falling object.

A more complete kinematic expression would be:
y = y_0 + v_0 t + 1/2 a t^2
 
yes but then the initial velocity can't be 2.9 but will be 0 so that we could use a=g
 
is Vo = initial velocity
and Y = vertical displacement
what does Yo = ? initial displacement.
 
kingyof2thejring said:
yes but then the initial velocity can't be 2.9 but will be 0 so that we could use a=g
Why can't the initial velocity be 2.9 m/s? a = -g.

Yes, y_0 is the initial displacement.
 
if we consider these two motions
one the piano being pulled up at 2.9ms-1 but the final velocity will become
0ms-1
two the piano falling down from the 0ms-1 to final velocity with -a therefore
-displacement taking vectors upwards to be positive.

or are we considering it as one motion where the displacement is 0
initial velocity is 2.9 but then how does a = -g.
 
You could consider the motion as one continuous segment (that's how I'd do it) or break it up into two segments (up and down); done correctly, both methods will give the same answer.

At the moment the rope breaks, the piano has an initial speed (2.9 m/s) and an initial height (6.0 m). Once the rope breaks the only force acting on the piano is gravity, so a = -g. The final height is zero.

You could break the motion up into two segments (up and down), but if you know how to use that kinematic equation there is no need to do that.
 
  • #10
0 = 6 + 2.9t - 4.9t^2
how do get t?
 
  • #11
if i use the quadratic formula "and should i be using it in the first place for such basic type of problem" but anyway i get 1.441 sec as +ve value
 
  • #12
Looks good to me.
 
  • #13
thank you__________
 
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