Question

A ball is dropped from a 45 metres high tower at the same time another ball is thrown upwards with a velocity of 40m/s. Find the relative speed of the balls.​

Answer 1

$\color{darkblue}\underline{\underline{\sf Given-}}$

• Height of tower (h) = 45m
• A ball is dropped from a tower whose initial Velocity is ${\sf (u_1)=0}$
• At the same time another ball is thrown with initial Velocity ${\sf (u_2)=40\:m/s}$

$\color{darkblue}\underline{\underline{\sf To \: Find-}}$

• Relative speed of the balls

Initial Speed of Ball 1 -

u = 0

Final Speed of Ball 1 -

$\color{orange}\implies{\sf v_1=-gt}$

Here ,

Negative sign is because velocity is in downward direction .

Initial Speed of Ball 2

$\color{orange}\implies{\sf u_2= +40m/s}$

Here , positive sign is because velocity direction is in upward Direction.

Final Speed of Ball 2 -

$\implies{\sf v_2=u_2-gt }$

$\color{orange}\implies{\sf v_2=(40-gt)}$

Relative Speed of Ball 1 wrt Ball 2

$\implies{\sf v_{12}=v_1-v_2 }$

$\implies{\sf v_{12}=-gt-(40-gt) }$

$\implies{\sf v_{12}=-gt-40+gt}$

$\color{red}\implies{\sf v_{12}=40m/s}$

$\color{darkblue}\underline{\underline{\sf Answer-}}$

Relative speed of balls is $\color{red}{\sf 40\:m/s}$.

Answer 2

Answer:

• The relative speed (v₁₂) of the balls is 40 m/s.

Given:

Ball - 1

• Initial velocity = 0 m/s.
• g = 10 m/s²

Ball-2

• Initial velocity = 40 m/s.
• g = - 10 m/s²

Explanation:

$\rule{300}{1.5}$

For Ball -1

Applying First Kinematic Equation.

⇒ v₁ = u₁ + a t

Where,

• v₁ Denotes Final velocity
• u₁ denotes Initial velocity.
• a Denotes Acceleration.
• t Denotes time period.

Now,

⇒ v₁ = u₁ + a t

Substituting the values,

⇒ v₁ = 0 + g × t

⇒ v₁ = g × t

⇒ v₁ = g t

⇒ v₁ = g t m/s².

∴ We got the velocity.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

For Ball -2

Applying First Kinematic Equation.

⇒ v₂ = u₂ + a t

Where,

• v₁ Denotes Final velocity
• u₂ denotes Initial velocity.
• a Denotes Acceleration.
• t Denotes time period.

Now,

⇒ v₂ = u₂ + a t

Substituting the values,

⇒ v₂ = 40 + (-g) × t

⇒ v₂ = 40 - g × t

⇒ v₂ = 40 - g t

⇒ v₂ = 40 - g t m/s².

∴ We got the velocity.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, Relative velocity (v₁₂)

As they are moving opposite to each other,

Therefore,

⇒ v₁₂ = v₁ + v₂

Substituting the values,

⇒ v₁₂ = g t + 40 - g t

⇒ v₁₂ = g t - g t + 40

⇒ v₁₂ = 40

⇒ v₁₂ = 40 m/s.

∴ The relative speed (v₁₂) of the balls is 40 m/s.

$\rule{300}{1.5}$