Question

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

Answer 1

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For the first object dropped from a height of 150m

U1 = 0

T1 = 2 s

S1 = ?

A1 = g

Let's use the 2nd equation of motion

$S = UT + \frac{1}{2}A{T}^{2}$

=》$0 + \frac{1}{2}\:G×{2}^{2}$

=》 2g

For the Second object dropped from a height of 100m

U2 = 0

T2 = 2 s

S2 = ?

A2 = g

Let's use the 2nd equation of motion

$S = UT + \frac{1}{2}\:A{T}^{2}$

=》$0 + \frac{1}{2}\:G×{2}^{2}$

=》 2 g

As the distance travelled by both objects is same so the difference in heights is equal to the initial separation = 50m

From the above, we observed that the distance travelled by both bodies during the fall remains the same for any time interval. So difference of height not change with time.

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Answer 2

Answer:

Explanation:

Solution :-

From 2nd equation of motion,

s = ut + 1/2at²

We know that,

g = 10 m/s

In 2 seconds,

Distance covered by first body, s₁ = 1/2 × g × 4

= 2g

= 2 × 10

= 20 m

Distance covered by second body, s₂ = 1/2 × g × 4

= 2g

= 2 × 10

= 20 m

After 2 seconds,

Height of 1st body, h₁ = 150 - 20 = 130 m

Height of 2nd body, h₂ = 100 - 20 = 80 m

Difference between height after 2 seconds, = 130 - 80 = 50 m.

Hence, The difference in heights of the objects will remain same with equal time as both the objects have been dropped from rest and are falling with same acceleration.