Question

When a body is dropped freely from a tower of height h, then the time taken by body to hit the ground is?

Answer 1

The time taken by the body to hit the ground is

                        $t= h/(u+1/2gt)$

If the initial velocity of a body is 'u', the time taken by the body to hit the ground is 't' and acceleration due to gravity is 'g'.

Then,

distance traveled by the body is calculated as,

                        $h = ut+1/2gt^2.....eq.(i)$

To find the time taken to hit the ground

                         h = t(u+1/2gt)

                         t = h/(u+1/2gt)

Answer 2

Answer:

Concept:

Free fall is defined as the moving mass in which gravitation is the only force acting on it. In general relativity, when gravity is reduced to a space-time curvature, a body in free fall has no force acting on it. Under standard Earth-bound conditions, a set of equations describing the motions of objects exposed to a constant gravitational pull. When a previously motionless object is allowed to fall freely under gravity, it travels a distance proportional to the square of the time elapsed.

Given:

When a body is thrown freely from a height of h, how long does it take for it to touch the ground?

Find:

find the time taken by body to hit the ground

Answer:

We can use the formula:

$S=u*t(\frac{1}{2})at^{2}$

that is the distance that the body must travel is

$S=H$

The remainder of the team has dropped the ball is

$u=0$

the acceleration of the body equivalent to gravity's acceleration is

$a=g$

the amount of time it takes for the body to reach the ground is

$t= T$

Then,

$u=0$

$H=(\frac{1}{2})*g*T^{2}$

$T=\sqrt{\frac{2H}{g} }$

∴ This is a broad formula that applies to any object that is dropped from height.

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