Question

A particle is thrown vertically upward from the surface of the earth.Let Tp be the time taken by the particle

Answer 1

Explanation:

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Answer 2

Given:

A particle is thrown vertically upwards from the surface of the earth.

Let TP be the time taken by the particle to travel from a point P above the earth to its highest point and back to the point P.

To find:

TQ is the time taken by the particle to travel from another point Q above the earth to its highest point and back to the same in terms of TP,TQ and H, is

Solution:

Using the standard symbols,

h = vi × t + 0.5 × g × t²

Since the object is falling, vi = 0.

rearranging the terms, t = √(2h/g)

So, we have obtained the equation for falling time,  

T = √(2h/g)

Using the attached diagram, we have,

Time taken from point P to point P = TP = 2√[2(h + H)/g]

⇒ TP² = 8(h + H)/g

∴ TP² = 8h/g + 8H/g .......(1)

Time taken from point Q to point Q = TQ = 2√[2h/g]

⇒ TQ² = 8h/g .......(2)

comparing equations (1) and (2), we get,

TP² = TQ² + 8H/g

8H/g = TP² - TQ²

g = 8H/[TP² - TQ²]

Hence the required equation.