Question

A body dropped from a tower covers 25m in the last second of its fall. Then, the height of tower is [ take g= 10 m/s^2]

(1) 45m (2) 20m (3) 40m (4) 50m

Answer 1

Distance travelled in the last second(nth second)=25m
Assuming u=0 and g=10 m/s^2
According to the formula,
S(nth)= u +a/2 (2n-1)
25 = 0+10/2 (2n-1)
25 = 5(2n-1)
25 = 10n -5
10n = 30
n=3 sec

That means, t=3 s
According to the second kinematical equation,
s = ut + 1/2 at^2
S = o +1/2 ×10× 9
S= 5 ×9
S = 45m
Hence, answer to this question is 45m