Q. A stone is dropped into water from a bridge 44.1m above the water. Another stone is thrown vertically downward 1s later. Both strike the water simultaneously. What was the initial speed of the second stone?
A) 12.25 $ms^{- 1}$
B) 14.75$\ ms^{- 1}$
C) 16.23$\ ms^{- 1}$
D) 17.15$\ ms^{- 1}$
Explanation
(a) $t = \sqrt{\frac{2 \times 44.1}{9.8}}$s=$\sqrt{9}$s =3s,
44.1$= v \times 2 + \frac{1}{2} \times 9.8 \times 2 \times 2$
or $2v = 44.1 - 4.9 \times 4 = 24.5$
or $v = \frac{24.5}{2}ms^{- 1} = 12.25ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.