Question

a particle is dropped from a tower. the distance covered by it in the last one second is equal to the distance covered in first 3 second. Find the height of Tower?

Answer 1

particle is drop in first 3 second

Answer 2

S=ut+ 1/2at2
S = 0×3 + 1/2 g (3×3) =45m
This is distance travelled in first three seconds
Now, distance travelled in last second is=
U + a/2 (2n-1) = 0 + g/2(2n-1)
This distance is equal to 45m
So, 10/2(2n-1) = 45
n=5
As 5 is the last second of the motion, so total time is 5 Seconds
t =
$\sqrt{2h \div g}$
25 = 2h / g
H = 125m
Height of the tower will be 125m
Hope it helps