Question

Please answer this ques from physics one D motion

Answer 1

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✤ Required Answer:

✒ GiveN:

• A ball is thrown downwards from a height of 120 m
• Another ball is thrown upwards with initial velocity of 40 m/s

✒ To FinD:

• After what time they will meet?

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✤ How to solve?

We need to know the equations of motion when a body is falling or going up in relation with Accleration due to gravity (g)

• v = u + gt
• s = ut + 1/2 gt²
• v² = u² + 2gs

Also, we need to remember some important points while answering this question,

• A body falling under the influence of g, has initial velocity = 0
• Accleration due to gravity has positive sign.
• And, A body going up have final velocity at highest point = 0
• Accleration due to gravity had negative sign$.$

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✤ Solution:

Let,

• Ball falling downwards be A
• And, ball going upwards is B
• Distance travelled by ball A be x
• Then, distance travelled by ball B = 120 - x

[ Because Ball A is falling from a height of 120m, and they meet at a point somwhere between this distance]

Ball A

• Initial velocity = 0 m/s
• Accleration due to gravity = 9.8 m/s²
• Distance travelled = x

By using 2nd equation of motion,

➝ s = ut + 1/2 gt²

➝ x = 0 × t + 1/2 × 9.8 × t²

➝ x = 4.9 t²........(1)

Ball B

• Initial velocity = 40 m/s
• Accleration due to gravity = -9.8 m/s²
• Distance travelled = 120 - x

By using 2nd equation of motion,

➝ s = ut + 1/2 gt²

➝ 120 - x = 40t + 1/2 (-9.8)t²

➝ 120 - x = 40t - 4.9t²........(2)

Substituting x in eq.(2),

➝ 120 - 4.9t² = 40t - 4.9t²

➝ 120 = 40t

➝ t = 3s

☀️ So, After 3 seconds, they will meet.

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