A rock falls from rest a vertical distance of 0.72 meter to the surface of a planet in .63 seconds. What is the magnitude of acceleration due to gravity on the planet?

Question

anonymous · Answer

the formula to be applied is : 
y = ut + .5a(t sqr)
u = 0 here. 
Try this, convey doubts if any.

anonymous · Answer

what does y, u , t stnd for?

anonymous · Answer

y is the displacement along the vertical direction ( which I have assumed as the y-axis  )
u = initial velocity of the particle
t is the time involved.

anonymous · Answer

and where does the .5 come from?

anonymous · Answer

ok, this is formula derived using some calculus. You want to know the derivation ?

anonymous · Answer

sure!

anonymous · Answer

wait either way i have to find acceleration not displacement

anonymous · Answer

$$dv = adt  $$Integrating this with proper limits ( i.e. v from initial velocity to final velocity and time similarly 0 to t )
The expression derived is $$v _{f} = v {_{i}} + at$$Next we use the expression derived above in$$dx = vdt$$Again integrating with proper limits we'll end up with the same formula I mentioned earlier.

anonymous · Answer

okay i understand it now thankyou! could you help with another question please? i'll post it in a sec!

anonymous · Answer

sure