Question

the highest of a tv transmission Tower at any place on the surface of the Earth is 245 metre the maximum distance of to which transmission Tower will be reach​

Answer 1

Answer:

Given:

Height of the TV transmission tower is 245 metres.

To find:

Max distance upto which the data can be transmitted.

Concept :

Here we will consider the signal waves to travel until it reaches in contact with the earth surface .

We call this "sight to sight" signal transmission.

Since height of the tower much less as compared to the earth radius, we can perform approximations . We apply Pythagoras theorem to get.

D max. = √(2gh)

Calculation:

Putting the respective values:

D max. = √(2gh)

=> D max. = √(2 × 10 × 245)

=> D max. = √4900

=> D max. = 70 metres.

Answer 2

Answer:

$\large\boxed{\sf{70\:m}}$

Explanation:

It's being given that the highest of a TV transmission tower at any place on the surface of Earth is 245 m.

To find the , Maximum distance up to which transmission will reach.

For Convenience, the signal waves are considered to travel until they reaches in contact with the earth surface .

• It's called "sight to sight" signal transmission.

But, height of the tower is much less as compared to the earth radius, thus, we can do approximations .

Also , We know that formula for maximum distance is given by,

$\large \boxed{ \sf{{D}_{max} = \sqrt{2gh} }}$

Where the symbols,

• h is the height of TV tower i.e., 245 m
• g is the acceleration due to gravity i.e., $\bold{10\:m{s}^{-2}}$.

Now, putting the respective values in the formula,

$\sf{= > {D}_{max} = \sqrt{2 \times 10 \times 245 } }\\ \\ \sf{= > {D}_{max} = \sqrt{490 \times 10}} \\ \\ \sf{= > {D}_{max} = \sqrt{4900} } \\ \\ \sf{ = > {D}_{max} = \sqrt{ {(70)}^{2} }} \\ \\ \sf{ = > {D}_{max} = 70 \: m}$

Hence, TV transmission tower will reach a maximum distance of 70 m.