Solution:
\[ 0.36 H= \frac{1}{2}g(t-1)^{2} \]
\[ H= \frac{1}{2}g(t)^{2} \]
Dividing we get
\[ 0.36 = \frac{(t-1)^{2}}{t^{2}} \]
\[ 0.6 = \frac{(t-1)}{t} \]
0.4t =1
t= 2.5 sec
S= (1/2) ×10 ×(2.5)²= 31.25 m
A body is dropped from a tower. It covers 64% distance of its total height in last second. Find out the height of tower [g = 10 ms–²]
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