A stone released with zero velocity from the top of a tower, reaches the ground in $$4\,s.$$ The height of the tower is $$\left( {g = 10\,m/{s^2}} \right)$$
A.
$$20\,m$$
B.
$$40\,m$$
C.
$$80\,m$$
D.
$$160\,m$$
Answer :
$$80\,m$$
Solution :
Initial velocity of stone $$u = 0$$
Time to reach at ground $$t = 4\,s$$
$$\eqalign{
& {\text{Acceleration }}a = + g = 10\,m/{s^2}\,\left( {{\text{As motion of body is along the acceleration due to gravity}}} \right) \cr
& \therefore {\text{Height of tower }}h = ut + \frac{1}{2}g{t^2} = \left( {0 \times 4} \right) + \frac{1}{2} \times 10 \times {4^2} = 80\,m \cr} $$
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