A) 95 m
B) 60 m
C) 80 m
D) 90 m
Correct Answer: C
Solution :
Let body reaches the ground in t seconds. \[\therefore \]Velocity of body after\[(t-2)\]seconds from equation of motion \[v=u+gt\] where v is final velocity, u the initial velocity (= 0) and \[t=t-2\] \[\therefore \] \[v=g(t-2)\] Distance covered in last two seconds \[h=g(t-2)\times 2+\frac{1}{2}g{{(2)}^{2}}\] \[60=20(t-2)+20\] On solving \[t=4\text{ }s\] Hence, height of tower is given by \[h=ut+\frac{1}{2}g{{t}^{2}}\] Since, \[u=0\] \[\therefore \] \[h=\frac{1}{2}g{{t}^{2}}\] \[=\frac{1}{2}\times 10\times {{(4)}^{2}}=80\,m\].You need to login to perform this action.
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