From a cliff 49 m high, a man drops a stone. One second later, he throws another stone. They both hit the ground at the same time. The speed of the second stone with which he threw:

A
$13.3 m / s$
B
$14.7 m / s$
C
$12.1 m / s$
D
$11.2 m / s$

Solution:

From a cliff

49 m

high, a man drops a stone. One second later, he throws another stone. They both hit the ground at the same time. The speed of the second stone with which he threw is

12.1 m / s

.
For first stone
Here initial speed

u = 0

, height

h = 49 m

, acceleration due to gravity,

g = 9.8 m / s^{2}

t = ?

Putting the values in the formula

h = ut + \frac{1}{2} g t^{2}

\Rightarrow 49 = 0 \times t + \frac{1}{2} \times 9.8 \times t^{2}

\Rightarrow 4.9 t^{2} = 49

\Rightarrow t^{2} = 10

\Rightarrow t = 3.16 s

For second stone

h = 49 m, g = 9.8 m / s^{2}, t = 3.16 - 1 = 2.16 s

Again

h = ut + \frac{1}{2} g t^{2}

\Rightarrow 49 = u \times 2.16 + \frac{1}{2} \times 9.8 \times {2.16}^{2}

\Rightarrow 2.16 u = 26.14

\Rightarrow u = 12.1 m / s

i.e. the second stone was thrown downward with speed of

12.1 m / s

From a cliff $49 m$ high, a man drops a stone. One second later, he throws another stone. They both hit the ground at the same time. The speed of the second stone with which he threw:

Solution:

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