A stone is dropped into water from a bridge 44.1 m above the water. Another stone is thrown vertically downward 1 sec later. Both strike the water simultaneously. What was the initial speed of the second stone

Question

A stone is dropped into water from a bridge 44.1 m above the water. Another stone is thrown vertically downward 1 sec later. Both strike the water simultaneously. What was the initial speed of the second stone (A) 12.25 / m s (B) 14.75 / m s (C) 16.23 / m s (D) 17.15 / m s

Tardigrade Admin · Accepted Answer

For the first stone t=√(2 h/g) =√(2 × 44.1/9.8)=3 Sec Time left for second stone =3-1=2 Sec Hence 44.1=u × 2+(1/2) 9.8(2)2 ⇒ 44.1-19.6=2 u ⇒ u=12.25 m / s

Q. A stone is dropped into water from a bridge $44.1 m$ above the water. Another stone is thrown vertically downward $1$ sec later. Both strike the water simultaneously. What was the initial speed of the second stone

12.25 / m s

14.75 / m s

16.23 / m s

17.15 / m s

For the first stone $t = \frac{2 h}{g}$
$= \frac{2 \times 44.1}{9.8} = 3$ Sec
Time left for second stone $= 3 - 1 = 2$ Sec
Hence $44.1 = u \times 2 + \frac{1}{2} 9.8 (2)^{2}$
$\Rightarrow 44.1 - 19.6 = 2 u$
$\Rightarrow u = 12.25 m / s$

Q. A stone is dropped into water from a bridge 44.1m above the water. Another stone is thrown vertically downward 1 sec later. Both strike the water simultaneously. What was the initial speed of the second stone