physics problem?

Question

a ball is dropped from the roof of a tower of height h.the total distance covered by it in the last sec of itz motion is equal to the distance covered in first three sec. What is the value of h in mts?
(g=10m/s^2)

on solving i got answer as 100mt but itz actual answer is 125mt

so plz help me

Answer 1

The distance covered in the first 3 seconds using
d = vt + a(t^2)/2
would be 45 meters, where v is the initial velocity. In the first 3 seconds that would be 0.

now using d = vt + a(t^2)/2 again set t to 1 second, a = 10, and d = 45 gives a v of 40 meters/second

then using v = at we can compute that the velocity was acheived after 4 seconds. Add the last second in and you get the object dropping for a total of 5 seconds. Then using our favorite d = vt + a(t^2)/2 with t = 5 and v = 0 gives 125 meters.

Answer 2

dist travelled in first 3 sec= 1/2 X A X t^2
= 1/2 X 10 X 9
= 45m
but 45m = dist. travelled in last sec = 1/2 A(2t - 1)
now t= 4
usind 2nd eq. we get s=80m
now total dist=80m + 45m
=125m

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