Physics, asked by Mister360, 1 month ago

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?

Answers

Answered by brainlychallenger99
2

Answer:

hey ,mate

Explanation:

→  Initial difference in height =(150−100)m=50m

→ We know that, by second equation of kinematics, s=ut+1 /2 at²

→ considering g=10m/s

→ Distance travelled by first body in

→2s=h1=0+(1 / 2)g(2)² =2g=2×10=20m

→ Distance travelled by another body in 2s=h2=0+

(1 / 2)g(2)²=2g=2×10=20m

→ After 2s, height at which the first body will be =h1=150−20=130m

→ After 2s, height at which the second body will be =h2=100−20=80m

→ Thus, after 2 s, difference in height =(130−80)

                =50m=initial difference in height.

→Thus, difference in height does not vary with time. So the answer is zero.

thank you

Answered by MяMαgıcıαη
7

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!

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Similar questions