Question

a particle is projected vertically up with velocity v=((4gr)÷3)^0.5 from earth surface. the velocity of particle at height equal to half of the maximum height reached by it​

Answer 1

Answer:

$v = \sqrt{\frac{gr}{3}}$

Explanation:

As we know that the projection velocity of the particle is given as

$v = \sqrt{\frac{4gr}{3}}$

when it reaches to its maximum height then the speed of the particle will be zero

So we will have

$KE_i + U_i = KE_f + U_f$

$\frac{1}{2}m(\frac{4gr}{3}) - mgr = 0 - \frac{mgr^2}{r + h}$

$- \frac{mgr}{3} = -\frac{mgr^2}{r + h}$

$r + h = 3r$

$h = 2r$

so we need to find the speed at half of the maximum height which is equal to H = r

so we will have

$KE_i + U_i = KE_f + U_f$

$\frac{1}{2}m(\frac{4gr}{3}) - mgr = \frac{1}{2}mv^2 - \frac{mgr^2}{r + r}$

$-\frac{mgr}{3} + \frac{mgr}{2} = \frac{1}{2}mv^2$

$\frac{mgr}{6} = \frac{1}{2}mv^2$

$v = \sqrt{\frac{gr}{3}}$

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Topic : Gravitational Potential energy

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Answer 2

Explanation:

Answer:

v = \sqrt{\frac{gr}{3}}v=

3

gr

Explanation:

As we know that the projection velocity of the particle is given as

v = \sqrt{\frac{4gr}{3}}v=

3

4gr

when it reaches to its maximum height then the speed of the particle will be zero

So we will have

KE_i + U_i = KE_f + U_fKE

i

+U

i

=KE

f

+U

f

\frac{1}{2}m(\frac{4gr}{3}) - mgr = 0 - \frac{mgr^2}{r + h}

2

1

m(

3

4gr

)−mgr=0−

r+h

mgr

2

- \frac{mgr}{3} = -\frac{mgr^2}{r + h}−

3

mgr

=−

r+h

mgr

2

r + h = 3rr+h=3r

h = 2rh=2r

so we need to find the speed at half of the maximum height which is equal to H = r

so we will have

KE_i + U_i = KE_f + U_fKE

i

+U

i

=KE

f

+U

f

\frac{1}{2}m(\frac{4gr}{3}) - mgr = \frac{1}{2}mv^2 - \frac{mgr^2}{r + r}

2

1

m(

3

4gr

)−mgr=

2

1

mv

2

−

r+r

mgr

2

-\frac{mgr}{3} + \frac{mgr}{2} = \frac{1}{2}mv^2−

3

mgr

+

2

mgr

=

2

1

mv

2

\frac{mgr}{6} = \frac{1}{2}mv^2

6

mgr

=

2

1

mv

2

v = \sqrt{\frac{gr}{3}}v=

3

gr