free falling bodies....solve this pls?

Question

if u drop an aspirin and it takes 0.18seconds to hit the table, how high above the table was the aspirin when it was released?

Answer 1

Here,
fall time, t = 0.18 second
gravitational acceleration, g = 9.8 m/sec^2
initial velocity, u = 0 m/s
[ when anything is 'dropped' it is implied that it's initial velocity was 0 ]
height, h = ?

We know,
h = ut + 1/2 gt^2
or, h = 1/2 gt^2 [ since u = 0 ]
or, h = 1/2 * 9.8 * (0.18)^2
so, h = 0.15876 meter

the height of the aspirin from the table was 0.15876 meter. (Ans.)

Hope this was helpful.

Answer 2

Assuming it was at rest initially (i.e. in someones hand)

s = ut + 1/2 a t^2

(ut is 0 because it was initially at rest) Therefore-:

s = 1/2 a t^2

(a = 9.81 - acceleration due to gravity)

1/2 x 9.81 x 0.18^2 = 0.158922 metres above the table

Answer 3

use s =ut + 1/2 at^2

S= Height above table
a =g = 9.8m/s^2
t = 0.18s
u = 0 as it was dropped

which gices us S = (1/2)*9.8 *(0.18)^2 0.31752m = (1/2)*31.75 cms

so height is 15.87 cm