Question

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Answer 1

Question

A ball which is thrown vertically upwards reaches the roof of a house 100 m high.At the moment this ball is thrown vertically upward,another ball is dropped from rest vertically downward from the roof of the house .At which height the balls will pass to each other

Solution

Find attachment for diagram ,

Lets find initial velocity with which ball thrown vertically upwards .

• Initial velocity , u = ? m/s
• Final velocity , v = 0 m/s

Finally attains rest

• Height , s = 100 m
• Acceleration , a = - 9.8 m/s²

Thrown against gravity

Apply 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ (0)² - u² = 2(-9.8)(100)

⇒ -u² = - 1960

⇒ u² = 1960

⇒ u = 44.27 m/s

So , Initial velocity = 44.27 m/s

Case - 1 : Ball 1

u = 44.27 m/s

s = x m

a = - 9.8 m/s²

Apply 2nd equation of motion ,

$\bf \to s=ut+\dfrac{1}{2}at^2\\\\\to \rm x=(44.27)t+\dfrac{1}{2}(-9.8)t^2\\\\\to \bf x=44.72t-4.9t^2...(1)$

Case - 2 : Ball 2

u = 0 m/s

Dropped from rest

s = ( 100 - x ) m

a = 9.8 m/s²

Apply 2nd equation of motion ,

$\bf s=ut+\dfrac{1}{2}at^2\\\\\to \rm (100-x)=(0)t+\dfrac{1}{2}(9.8)t^2\\\\\to \bf 100-x=4.9t^2...(2)$

Add (1) and (2) ,

$\to \rm x+100-x=44.27t-4.9t^2+4.9t^2\\\\\to \rm 100=44.27t\\\\\to \bf t=2.26\ s$

Now , sub. t value in (2) ,

$\to \rm 100-x=4.9(2.26)^2\\\\\to \rm 100-x=25.02\\\\\to \bf x=74.98\\\\\to \bf x\cong 75\ m$

So , both will pass each other at x = 75 m

Answer 2

Answer:

Question

A ball which is thrown vertically upwards reaches the roof of a house 100 m high.At the moment this ball is thrown vertically upward,another ball is dropped from rest vertically downward from the roof of the house .At which height the balls will pass to each other

Solution

Find attachment for diagram ,

Lets find initial velocity with which ball thrown vertically upwards .

Initial velocity , u = ? m/s

Final velocity , v = 0 m/s

Finally attains rest

Height , s = 100 m

Acceleration , a = - 9.8 m/s²

Thrown against gravity

Apply 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ (0)² - u² = 2(-9.8)(100)

⇒ -u² = - 1960

⇒ u² = 1960

⇒ u = 44.27 m/s

So , Initial velocity = 44.27 m/s

Case - 1 : Ball 1

u = 44.27 m/s

s = x m

a = - 9.8 m/s²

Apply 2nd equation of motion ,

$\begin{gathered}\bf \to s=ut+\dfrac{1}{2}at^2\\\\\to \rm x=(44.27)t+\dfrac{1}{2}(-9.8)t^2\\\\\to \bf x=44.72t-4.9t^2...(1)\end{gathered} </p><p>$

Case - 2 : Ball 2

u = 0 m/s

Dropped from rest

s = ( 100 - x ) m

a = 9.8 m/s²

Apply 2nd equation of motion ,

$\begin{gathered}\bf s=ut+\dfrac{1}{2}at^2\\\\\to \rm (100-x)=(0)t+\dfrac{1}{2}(9.8)t^2\\\\\to \bf 100-x=4.9t^2...(2)\end{gathered} </p><p>$

Add (1) and (2) ,

$\begin{gathered}\to \rm x+100-x=44.27t-4.9t^2+4.9t^2\\\\\to \rm 100=44.27t\\\\\to \bf t=2.26\ s\end{gathered} </p><p>$

Now , sub. t value in (2) ,

$\begin{gathered}\to \rm \pink{100} \red{-x=4.9}(2.26)^2\\\\\to \rm 100-x=25.02\\\\\to \bf x=74.98\\\\\to \bf x\cong 75\ m\end{gathered} </p><p>$

So , both will pass each other at x = 75 m