Question

Two bodies fall freely from different heights and reach the ground simultaneously.the time of descent for the first body is 2 s and for the second body is 1 s.at what height was the first body situated when the other began to fall?

Answer 1

Given:

Two bodies fall freely from different heights and reach the ground simultaneously.

The time of descent for the first body is 2 s and for the second body is 1 s.

To find:

At what height was the first body situated when the other began to fall?

Solution:

We use the formula,  

S = ut + 1/2 at²    

(equation of distance travelled with respect to time)

where S = distance  

u  = initial velocity

a = acceleration

t =  time

Let the first body be at a height of (h + x) distance.

Let the second body be at a height of (h) distance.

So, we have,

S1 = 0 + 1/2 × 9.8 × 2²

⇒ h + x = 19.6 ....(1)

S2 = 0 + 1/2 × 9.8 × 1²

⇒ h = 4.9 .......(2)

using equation (2) in (1), we get,

4.9 + x = 19.6

x = 19.6 - 4.9

∴ x = 14.7 m

At a height of 14.7 m the first body situated when the other began to fall.

Answer 2

Answer:

Explanation:

Let h1 and h2 be the respective heights of two bodies. Here both the bodies are allowed to fall freely. If t1 and t2 are their time of descent then,

h1 = ut + 1/2 gt²

So using this equation we can find h1 and h2

h1 = 1/2×9.8×2²

= 9.8×2

= 19.6 m

h2 = 1/2×9.8×1²

= 4.9 m

h1-h2= 19.6 - 4.9 = 14.7 m