# Time taken for body under free fall

Swetasuria

## 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

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

Looks wrong to me. Time can't be negative, can it? Also, shouldn't the answer be less than 3?

Mentor
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?

Swetasuria
If the origin is on the ground, the distance (which is above the origin) will be positive, right?

Mentor
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.

Swetasuria
Okay. So y=yo+ut-gt2/2
0=45-30t-5t2
t=1.242s

Mentor
Okay. So y=yo+ut-gt2/2
0=45-30t-5t2
t=1.242s
Looks good.