Question

A body falls freely without initial velocity from a height of 20meters at what height will it reach a velocity equal to half the velocity with which it reaches the ground

Answer 1

h = ut + 1/2gt2

20 = 0 + 1/2*10*t2

t = 2 sec

v = u + gt

v = 0 + 10*2

v = 20

at half velocity v = u + gt

10 = 0 + 10 /t

t = 1sec

at half velocity, s = ut + 1/2gt2

s = 0 + 1/2*10*1

s = 5 m from top

s = 15 m from ground

Answer 2

Answer:

At 15 meters of height, the body will reach a velocity equal to half the velocity with which it reaches the ground.

Explanation:

The velocity after reaching the ground when the body is released from height "h" is given as,

$\mathrm{v=\sqrt{2gh} }$             (1)

Where,

v=velocity at 20 m of height

g=acceleration due to gravity=10m/s²

h=height from where the body is released

From the question, we have that the height from where the body is released is 20 meters. So, the velocity at this point is,

$\mathrm{v=\sqrt{2\times 10\times 20} }$

$\mathrm{v=20\ m/s}$              (2)

Now for the rest part of the question, we will use the first equation of motion which is given as,

$\mathrm{v^2-u^2=2gs}$              (3)

Where,

v=final velocity

u=initial velocity

g=acceleration due to gravity=10 m/s²

s=the required height

So,

The final velocity(v)=20 m/s

The initial velocity(u)=10 m/s                 (velocity equal to half the velocity with which it reaches the ground)

By inserting the values in equation (3) we get;

$\mathrm{20^2-10^2=2\times 10\times s}$

$\mathrm{400-100=2\times 10\times s}$

$\mathrm{300=2\times 10\times s}$

$\mathrm{s=\frac{300}{20} }$

$\mathrm{s=15 \ m}$

Hence, at 15 meters of height, the body will reach a velocity equal to half the velocity with which it reaches the ground.

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