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
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
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 ::
- 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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