Question

A boy on a cliff 49 m high drops a stone. One second later he throws a second stone after the first. They both hit the ground at the same time. With what speed did he throw the second stone?​

Answer 1

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$\huge\mathfrak{Answer:-}$

⏩ i) For the first stone:

u= 0, s= -49 m, g= -9.8 m/s²

As, s= ut + 1/2gt²

Therefore,

-49= 0-1/2 × 9.8 × t²

or

t= √49× 2/9.8

= √10 = 3.16 s.

⏩ ii) Time taken by the second stone to hit the ground= 3.16-1 = 2.16 s

Now, t= 2.16 s, s= -49 m, g= -9.8 m/s²

As,

s= ut + 1/2gt²

Therefore,

-49 = u × 2.16 - 1/2 × 9.8 × (2.16)²

or

-49 = 2.16 u - 22.86

or

-26.14 = 2.16 u

or

u= - 26.14/2.16

= -10.1 m/s → The speed acquired by the second stone. (Ans.)

[The negative sign shows that the velocity is in the downward direction]

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Answer 2

Explanation:

Answer:−

⏩ i) For the first stone:

u= 0, s= -49 m, g= -9.8 m/s²

As, s= ut + 1/2gt²

Therefore,

-49= 0-1/2 × 9.8 × t²

or

t= √49× 2/9.8

= √10 = 3.16 s.

⏩ ii) Time taken by the second stone to hit the ground= 3.16-1 = 2.16 s

Now, t= 2.16 s, s= -49 m, g= -9.8 m/s²

As,

s= ut + 1/2gt²

Therefore,

-49 = u × 2.16 - 1/2 × 9.8 × (2.16)²

or

-49 = 2.16 u - 22.86

or

-26.14 = 2.16 u

or

u= - 26.14/2.16

= -10.1 m/s → The speed acquired by the second stone. (Ans.)

[The negative sign shows that the velocity is in the downward direction]