Question

a body is droped from a heigt of 250m and semutaneously another body is droped from a height of 200m what is a diffrencein height after they have fallen for 5 seconds{g=9.8ms-2}

Answer 1

Answer:

The difference in height after they have fallen for 5 seconds is 50m.

Explanation:

Given,

A body is dropped from a height of 250m and simultaneously another body is dropped from a height of 200m.

∴ Initially, the difference between their heights is = 250 - 200 = 50m.

Now, we've to calculate the difference between their heights after they've fallen for 5 sec.

Let's calculate the distance covered by the first body in 5 sec.

We know that,

→ s = ut + ½.g.t²

Where,

• u(intial velocity) = 0 [Since it was at rest]
• t(time) = 5sec.
• g(acceleration due to gravity) = 9.8m/s²

Substitute the values,

→ s = 0(5) + ½.(9.8)(5)²

→ s = 0 + ½.(9.8)(25)

→ s = 122.5m

∴ Remaining distance = 250 - 122.5 = 127.5m

Similarly, for the second object :-

→ s = ut + ½.a.t²

→ s = 0 + ½.(9.8)(25)

→ s = 122.5m

∴ Remaining distance = 200 - 122.5 = 77.5m

Hence, after falling for 5 seconds the distance between them would be :-

= 127.5 - 77.5

= 50m

∴ Required answer:- 50m

__________________________

Formula used,

• s = ut + ½at²

Where,

• s denotes the distance.
• u = intial velocity.
• v = final velocity.
• t = time taken.

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Answer 2

☀️ Given that: A body is dropped from a height of 250m and simultaneously another body is dropped from a height of 200m

      Need to find: Difference of height after they have fallen for 5 seconds ( g = 9.8 m/s² )

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Must Know:

• Initial velocity of both the bodies will be taken as zero as they are initially at rest before being dropped.

• Acceleration acting upon both the bodies will be positive gravitational acceleration ( 9.8 m/s² ) as it is dropped down [ Note : The gravitational acceleration would be negative if they were thrown upwards ]

• According to Second Equation of motion s =  ut + ½at²

Where,

s is distance covered

u is initial velocity ( 0 m/s )

t is time taken ( 5 m/s )

a is acceleration ( 9.8 m/s² )

Solution:

• For First Body -

⇒ s = 0 × 5 +  ½ × 9.8 × ( 5 )²

⇒ s = 0 +  ½ × 9.8 × 25

⇒ s = 245/2

⇒ s = 122.5m

As it was dropped from height of 250m,

∴ Distance from ground = 250 - 122.5

= 127.5m

• For another body -

⇒ s = 0 × 5 +  ½ × 9.8 × ( 5 )²

⇒ s = 0 +  ½ × 9.8 × 25

⇒ s = 245/2

⇒ s = 122.5m

As it was dropped from height of 200m,

∴ Distance from ground = 200 - 122.5

= 77.5m

Now, we know their distance from ground and we can find the difference in their heights -

→ 127.5 - 77.5

→ 50m

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• Henceforth, the distance in their heights after they have fallen from 5 seconds is 50m