Time taken for body under free fall

AI Thread Summary
A ball dropped from a cliff takes 3 seconds to fall halfway, leading to a calculated height of 90 meters. The discussion highlights the importance of consistent sign conventions in kinematic equations, particularly when determining displacement and time. A sign error was identified in the initial calculations, prompting a clarification of the position function. After correcting the approach, the time taken for the second half of the descent was found to be approximately 1.242 seconds. The conversation emphasizes the need for careful application of physics principles in solving motion problems.
Swetasuria
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Homework Statement


A ball is dropped from the top of a cliff. If time taken for half of the descent is 3s, then what is the time taken for the rest of rest of the descent? Take g=10m/s2


Homework Equations


T=√(2h/g)
v=u-gt
s=ut-gt2/2

The Attempt at a Solution


T=√(2h/g)
h=T2g/2=45m

∴height of cliff=90m

v=u-gt=0-(10*3)=-30ms-1

s=ut-gt2/2
45=-30t-5t2
t=-3

:frown: Looks wrong to me. Time can't be negative, can it? Also, shouldn't the answer be less than 3?
 
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Swetasuria said:
s=ut-gt2/2
45=-30t-5t2
t=-3
You're making a sign error. Use the sign convention consistently. If down is negative, what should be the sign of s?
 
If the origin is on the ground, the distance (which is above the origin) will be positive, right?
 
Swetasuria said:
If the origin is on the ground, the distance (which is above the origin) will be positive, right?
No. What you're calling 's' is really Δs, which is negative as the object is falling.

The full statement of position as a function of time would be (using y instead of s, for clarity):
y = y0 + ut - 1/2gt2

In your case, y = 0 (the final position, measured from the ground) and y0 = 45.

Let me know if that's clear.
 
Okay. So y=yo+ut-gt2/2
0=45-30t-5t2
t=1.242s
 
Swetasuria said:
Okay. So y=yo+ut-gt2/2
0=45-30t-5t2
t=1.242s
Looks good.
 
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