Motion in a Straight Line - Result Question 45
47. A stone released with zero velocity from the top of a tower, reaches the ground in $4 sec$. The height of the tower is $(g=10 m / s^{2}) \quad$ [1995]
(a) $20 m$
(b) $40 m$
(c) $80 m$
(d) $160 m$
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Answer:
Correct Answer: 47. (c)
Solution:
- (c) Initial velocity $(u)=0$; Time $(t)=4 sec$ and gravitational acceleration $(g)=10 m / s^{2}$.
Height of tower
$h=u t+\frac{1}{2} g t^{2}=(0 \times 4)+\frac{1}{2} \times 10 \times(4)^{2} .=80 m$.
