Question

Hello Guys!! Good morning. Can anyone solve this question,it's urgent!!​

Answer 1

Given:-

For first Stone

• Height of Cliff = 78.4m

• Acceleration Due to Gravity = +9.8m/s²

• Initial Velocity = 0m/s (As it comes to rest)

For Second Stone:-

• Height of Cliff = 78.4m

• Acceleration Due to Gravity = +9.8m/s²

To Find:-

• Initial Velocity with which Second Stone was thrown.

Formulae Used :-

• S = ut + ½ × a × t²

Now,

Time take by first stone to reach ground.

S = ut + ½ × a × t²

78.4 = 0 × t + ½ × 9.8 × t²

78.4 = 0 + 4.9t²

t² = 78.4/4.9

√t² = √16

t = 4s.

Hence, The time taken by first stone to reach ground is 4 second.

Or, Atq.

Time taken by second stone = 4 - 1 = 3s.

Now,

Speed with which Second Stone was thrown:-

S = ut + ½ × a × t²

78.4 = u × 3 + ½ × 9.8 × 3²

78.4 = 3u + 4.9 × 9

78.4 = 3u + 44.1

78.4 - 44.1 = 3u

34.3 = 3u

u = 34.3/3

u = 11.43m/s

Hence, The speed of the second stone was 11.43m/s

Answer 2

Answer:

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