A mountaineer standing on the edge of a cliff 441 m above the ground throws a stone horizontally with an initial speed of 20 m/s-1 . What is the speed with which the stone reaches the ground?

Question

A mountaineer standing on the edge of a cliff 441 m above the ground throws a stone horizontally with an initial speed of 20 m/s-1 . What is the speed with which the stone reaches the ground? (A)  90 m/s-1  (B)  95.08 m/s-1  (C)  85 m/s-1  (D)  92 m/s-1

Tardigrade Admin · Accepted Answer

⇒ Here Ux will remain constant. ⇒ Vy2=Uy 2+2 ay Sy Vy2=0+2 × 10 × 441 Vy=√8820 Vy=94.25 m / s -1 So, let U=√Vx2+Vy2 =√(20)2+(94.25)2 =√400+8820=√9220 =96 ms -1

Q. A mountaineer standing on the edge of a cliff $441 m$ above the ground throws a stone horizontally with an initial speed of $20 m / s^{- 1}$ . What is the speed with which the stone reaches the ground?

$90 m / s^{- 1}$

$95.08 m / s^{- 1}$

$85 m / s^{- 1}$

$92 m / s^{- 1}$

$\Rightarrow$ Here $U_{x}$ will remain constant.
$\Rightarrow V_{y}^{2} = U_{y}^{2} + 2 a_{y} S_{y}$
$V_{y}^{2} = 0 + 2 \times 10 \times 441$
$V_{y} = 8820$
$V_{y} = 94.25 m / s^{- 1}$
So, let $U = V_{x}^{2} + V_{y}^{2}$
$= (20)^{2} + (94.25)^{2}$
$= 400 + 8820 = 9220$
$= 96 m s^{- 1}$

Q. A mountaineer standing on the edge of a cliff 441m above the ground throws a stone horizontally with an initial speed of 20m/s−1 . What is the speed with which the stone reaches the ground?

90m/s−1

95.08m/s−1

85m/s−1

92m/s−1