Question

A stone is dropped from the top of a tower of height 40m. The stone falls from rest and air resistance is negligible. What is the time taken for it to fall the last 10m to the ground?

Answer 1

Ok I assumed:
a=9.81
u=0
s=10m

And using the equation s= ut + 1/2 at2 I calculated the time to be 2.85--2.9 seconds.
Buuuuut my answer is wrong!! Did i assume something wrong?

Answer 2

Use S=Ut+1/2at^2
40=0t+1/2*9.81t^2
40=4.905t^2
8.15=t^2
t=2.86

that is the total time for the stone to fall 40m
now if we divide the time by 4 since 40/4=10m
we get a time of .71 seconds

Answer 3

does this make sense at all?

Answer 4

It does, but the answer is 0.38s !!
Idk maybe i'm mindblocked, idk how to get thisssssssss. :/

Answer 5

ok i think i got it!
if u figure out how long it take to fall 30m and figure out how long it takes for the stone to fall 40m all you have to do is subtract the 40m time- 30m time and you will get .38s

Answer 6

2.86-2.47=.38s

Answer 7

Vi=0
H=40m
h=H-10m=30m
g=9.8m/*s^2
As we know that
2gh=Vf^2-Vi^2
2*9.8*30=Vf^2
Vf=24.2487m/s
now as
ds/dt=Vf=24.2487
ds=24.2487dt
integratin between
\[0\le s \le10m \]
we get
10m=24.2487t
t=10/24.2487=0.40 sec

Answer 8

@Brent0423 ohhhhhhhhhhhhhhhhhhhhhhhhhh!!! Gotcha. Thanks a bunch!