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

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

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

 

[Doc Al]

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?

 

[Doc Al]

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=y_o+ut-gt²/2
0=45-30t-5t²
t=1.242s

 

[Doc Al]

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

Looks good.