A) 50m
B) 75m
C) 100m
D) 125m
Correct Answer: D
Solution :
Let height of tower is h and body takes time t to reach the ground when it falls freely. \[\therefore \] \[h=\frac{1}{2}g{{t}^{2}}\] ...(i) In last second ie, tth sec body travels \[=0.36h\] It means in rest of the time ie, in \[(t-1)sec\]it travels \[=h-0.36\text{ }h=0.64\text{ }h\] Now, applying equation of motion for \[(t-1)sec\] \[0.64\,h=\frac{1}{2}g{{(t-1)}^{2}}\] ....(ii) From Eqs. (i) and (ii), we get \[t=5s\]and \[h=125m\]
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