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 acceleration? How does the difference in
heights vary with time?
Answers
Answer:
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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 2s :
↦ Height of 1st object = 150 - 20
➦ Height of 1st object = 130 m
❏ Calculating height of second object after 2s :
↦ 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 2s = 130 - 80 = 50 m. So, both differences are same.
∴ Hence, difference in their height does not vary with time!