Question

ANSWER CORRECT AND FAST. BE BRAINLIEST!!!!



A ball is thrown from a height of 200m with velocity of 5m/s. Another ball is thrown , at the same instant, upwards from ground with velocity 20m/s.
When and where will the balls meet??

NO SPAMMING.


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Answer 1

Answer:-

• The balls will meet at$\large\leadsto\boxed{\tt\purple{8 \: seconds}}$

• They will meet at$\large\leadsto\boxed{\tt\purple{353.6 \: m}}$

• Given:-

• Ball is dropped from a height of 200 m with a velocity of 5 m/s

• Initial velocity of second ball, u₁ = 5 m/s

• Acceleration , a = 9.8 m/s²

Also,

• Ball is thrown upward with a velocity of 20 m/s

• Initial velocity of first ball,u₂ = 20 m/s

• Acceleration a = -9.8 m/s² [ Negative as the ball is going upwards ]

• To Find:-

• When and where the two balls will meet

• Solution:-

Let, first and second ball meet after a time 't' and covering distance of s₁ and s₂ .

★ For 1st ball thrown downward:-

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• Using second equation of motion

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➪ $\sf s_1 = u_1 t + \dfrac{1}{2}a_1 t^{2}$

➪ $\sf s_1 = 5 t + \dfrac{1}{2} (9.8) t^2$

➪ $\sf s_1 = 5 t + 4.9 t^2 \dashrightarrow\bf\red{[eqn.i]}$

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Now,

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★ For 2nd ball thrown upward:-

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• Using second equation of motion:-

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➪ $\sf s_2 = u_2 t + \dfrac{1}{2} a_2 t^{2}$

➪ $\sf s_2 = 20 t + \dfrac{1}{2} (-9.8) t^2$

➪ $\sf s_2 = 20 t - 4.9 t^2 \dashrightarrow\bf\red{[eqn.ii]}$

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Since, one ball is thrown from ground and another from a 200 m building,

Hence,

➪ $\sf s_1 + s_2 = 200 m$

➪ $\sf 5 t + 4.9 t^2 + 20 t - 4.9 t^2 = 200$

➪ $\sf 25 t = 200$

➪$\sf t = \dfrac{200}{25}$

★ $\bf\pink{t = 8 \: seconds}$

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• Substituting value of t in eqn[i]:-

➪ $\sf<strong> </strong>s_2 = 5 t + 4.9 t^2$

➪ $\sf s_2 = 5 \times 8 + 4.9 (8)^2$

➪ $\sf s_2 = 40 + 4.9 \times 64$

➪ $\sf s_2 = 40 + 313.6$

➪ $\sf s_2 = 353.6$

★ $\bf\pink{s_2 = 353.6 \: metres}$

Therefore,

Two balls will meet after a time 8 seconds.

They will meet at a distance of 353.6 metres above the ground.

★ Note:- Height of 353.6 m is not possible. Hence, the given data is inappropriate.

Answer 2

Given:

• Height = 200m
• Initial Velocity of 1st Ball = u1= 5m/s
• Initial Velocity of 2nd Ball= u2= 20m/s

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Need to find:

• When and where the balls will meet?

━━━━━━━━━━━━━━━━

Solution:

For the second ball (which is projected upwards),

★ Maximum height :

= $\dfrac{ {u}^{2} }{2g}$

$= \dfrac{ {20}^{2} }{2 \times 10} m$

$= \dfrac{400}{20} m$

$\bold{\boxed{\red{ = 20m}}}$

★ Time taken by the ball to reach maximum height:

$t = \dfrac{u}{g}$

$\implies \: t = \dfrac{20}{10}$

$\bold{\boxed{\purple{ \implies \: t = 2sec}}}$

Thus, at 2sec the ball will reach its maximum height. And after 2secs the 2nd Ball will start falling down again.

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Now for the 1st Ball (which is dropped from tower)

★ In 2secs it'll cover:

$s = ut + \dfrac{1}{2} a {t}^{2}$

$s = 5 \times 2 + \dfrac{1}{2} \times 10 \times {2}^{2}$

$\implies \: s = 10 + 5 \times 4m$

$\bold{\boxed{\red{ \implies \: s = 30 \: m}}}$

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Thus in 2sec the 1st Ball will cover 30m from top.

It will be at a height:

→ 200- 30m from ground

→ 170m from ground

At 2sec 2nd ball will be at a height of 20m from ground.

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→ Thus clearly, the two balls will never meet at a height above the ground.

→ However the two balls will meet on the ground!