Question

Answer This; Mods And Stars!

An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time?

$\blue{ \boxed{ \sf \color{cyan}\: Note: Don't \: Spam}}$
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Answer 1

⇝ Given :-

• An object is dropped from rest from a height of 150 m .

• Simultaneously another object is dropped from rest from a height of 100 m.

⇝ To Find :-

• Difference b/w heights after 2 s.

• How does the difference in heights vary with time ?

⇝ Solution :-

Here Acceleration is the Acceleration due to gravity (g).

• Acceleration = a = g = 10 m/s²

❒ We Have 2nd Equation of Kinematics as :

$\red \bigstar \: \boxed{\orange{ \boxed{ \bf{s =ut} + \dfrac{1}{2}{at}^{2}}}}$

★ For First Body :-

• Time = t = 2 second

• Initial Velocity = u = 0 m/s

• Acceleration = a = 10 m/s²

❒ Using 2nd Equation of Kinematics :

★ Distance covered by 1st body :

$\text s_1 = 0 + \frac{1}{2} \times 10 \times {2}^{2}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf s_1 = 20 \: m} }}}$

Hence,

Height of 1st body after 2s = 150 - 20

⇝Height of 1st body after 2s = 130m

Now,

★ For Second Body :-

• Time = t = 2 second

• Initial Velocity = u = 0 m/s

• Acceleration = a = 10 m/s²

❒ Using 2nd Equation of Kinematics :

★ Distance covered by 2nd body :

$\text s_2 = 0 + \frac{1}{2} \times 10 \times {2}^{2}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf s_2 = 20 \: m} }}}$

As Initial Height Second body is 100 m

Hence,

Height of 2nd body after 2s = 100 - 20

⇝Height of 2nd body after 2s = 80m

So,

Difference in their Heights after 2s = 130 - 80

$\underline{\bf \pink{\underline{Difference \: in \: their \: Heights}}}\\ \underline{\bf \pink{\underline{after \: 2s = 50 m }}}$

Also,

It concluded that initial difference between their Heights is equals to final difference between their Heights

So,

$\underline{\bf \pink{\underline{Difference\: in\: their\: Heights}}}\\ \underline{\bf \pink{\underline{ does\: not\: vary\: with\: time}}}$

Answer 2

Question: $\:$

• An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time?

Answer:

• Difference between their heights after 2 s is 50 m.
• Difference in their height does not vary with time.

Step By Step Explanation:

Given that:

• An object is dropped from rest at a height of 150 m
• Simultaneously another object is dropped from rest at a height 100 m

To Find:

• What is the difference in their heights after 2 s if both the objects drop with same accelerations?
• How does the difference in heights vary with time?

Solution:

• Firstly let's calculate distance covered by first object in 2 s. Using second equation of motion ::
• $\LARGE\pmb{\boxed{\bf{s = ut + {}^{1}/{}_{2}\:gt^2}}}$
• Where, s denotes distance covered, u denotes initial velocity, t denotes time taken, g denotes acceleration due to gravity. We have, u = 0 m/s, g = 10 m/s² and t = 2 s.

★ Putting all values in formula :

↪ s = (0 × 2) + ½ × 10 × (2)²

↪ s = 0 + ½ × 2 × 2 × 10

↪ s = 1 × 2 × 10

↪ s = 2 × 10

➢ s = 20 m

∴ Hence, distance covered by first object in 2 s is 20 m.

• Now, let's calculate the distance covered by second object in 2 s by using same formula. Here, we have u = 0 m/s, g = 10 m/s² and t = 2 s.

★ Putting all values in formula :

↪ s = (0 × 2) + ½ × 10 × (2)²

↪ s = 0 + ½ × 2 × 2 × 10

↪ s = 1 × 2 × 10

↪ s = 2 × 10

➢ s = 20 m

∴ Hence, distance covered by second object in 2 s is 20 m.

• Now, let's calculate heights of both objects after 2s ::

❏ Calculating height of first object after 2 s :

↦ Height of 1st object = 150 - 20

➦ Height of 1st object = 130 m

❏ Calculating height of second object after 2 s :

↦ Height of 2nd object = 100 - 20

➦ Height of 2nd object = 80 m

∴ Hence, heights of both objects after 2 s are 130 m and 80m.

• Now, let's find difference between their heights, We can easily find it by subtracting height of second object after 2 s from height of first object after 2 s.

❏ Finding difference b/w their heights :

⇒ Difference b/w their heights = 130 - 80

➠ Difference b/w their heights = 50 m

∴ Hence, difference between their heights after 2 s is 50 m.

ㅤㅤㅤㅤㅤ— — — — — — — — — —

We know that initial difference between their heights = 100 - 50 = 50 m and difference between their heights after 2 s = 130 - 80 = 50 m. So, both differences are same.

∴ Hence, difference in their height does not vary with time!

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