Question

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

Answer 1

Explanation:

EXPLANATION.

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.

Find out the speed with which he threw

the second stone.

$\sf : \implies \: for \: first \: stone \: \\ \\ \sf : \implies \: s \: = 49 \: m \: \\ \\ \sf : \implies \: g \: = 9.8ms {}^{ - 2} \\ \\ \sf : \implies \: \green{{ \underline{from \: second \: equation \: of \: kinematics}}} \\ \\ \sf : \implies \: s \: = ut \: + \: \frac{1}{2}a {t}^{2}$

$\sf : \implies \: 49 = 0 \: + \: \dfrac{1}{2} \times 9.8 \times {t}^{2} \\ \\ \sf : \implies \: {t}^{2} = \frac{49 \times 2}{9.8} \\ \\ \sf : \implies \: t \: = \sqrt{10} = 3.16 \: seconds$

$\sf : \implies \: for \: second \: stone \: \\ \\ \sf : \implies \: t \: = \sqrt{10} - 1 \: seconds \\ \\ \sf : \implies \: \: \green{{ \underline{from \: newton \: 2 {}^{nd} equation \: of \: \: kinematics}}} \\ \\ \sf : \implies \: s = ut \: + \: \frac{1}{2} a {t}^{2}$

$\sf : \implies \: 49 = u \: \times \sqrt{10} - 1 \: + \: \dfrac{1}{2} \times 9.8 \times ( \sqrt{10} - 1) {}^{2} \\ \\ \sf : \implies \: 49 = 2.16u \: + \: \frac{1}{2} \times 9.8 \times 2.16 \times 2.16 \\ \\ \sf : \implies \: 49 = 2.16u \: + \: 22.8614 \\ \\ \sf : \implies \: u \: = 12.101 \: m {s}^{1}$

$\sf : \implies \: \orange{{ \underline{the \: speed \: with \: which \: he \: throw \: the \: second \: stone \: = 12.101 \: ms {}^{ - 1} }}}$

Answer 2

for first stone 

h = 49m

g = 9.8m/s²

t = ?

u = 0

apply 

h = ut + gt²/2

49 = 0 + 9.8t²/2

t² = 10

t = √10 = 3.16s

for second stone

t = 3.16 - 1 = 2.16s

h = 49m

g = 9.8m/s²

u = ?

again apply

h = ut + gt²/2

49  = u*2.16 + 9.8*2.16²/2

49 = 2.16*u + 22.86

26.14/2.16 = u

u = 12.10m/s