Question

guys plss answer my question in detail​

Answer 1

Answer:

√2u

Explanation:

Given that if body is project at velocity of 'u' then it attains the height 'h'.

We know that when the body reaches at height , the maximum point in upward direction ; it's velocity will be zero at this point.

Thus in this case we have,

Initial Velocity = u

Final Velocity (v) = 0

Acceleration = g

Distance travelled(s) = h

Now by using third equation of motion,

v² = u² - 2gs

0 = u² - 2gh

h = u²/2g ..................I

Now,

If we want it to attain double height that is '2h' Let that we projected it with velocity u' .

Then,

Distance travelled (s) = 2h

Final Velocity (v) = 0

Then Again using third equation of motion,

v² = u² - 2gs

0 = u'² - 2g(2h)

u'² = 4gh

Putting the value of h from equation I)

u'² = 4 × (u²/2g)

u'² = 2u²

u' = u√2

This Option C)√2u is correct

Answer 2

Given:-

A body is attaining height h with a velocity u.

Height = h m

Final Velocity = v = 0 m/s

Initial Velocity = u m/s

Acceleration due to Gravity = g m/s²

Now, In first case using third equation of Motion.

v² = u² - 2gh

0² = u² - 2gh

h = u²/2g -----------(1)

Now, When height is doubled,means h = 2h and the velocity, u = u'

v² = u'² -2gh

0² = u'² -2gh

u'² = 2gh

Putting the values of 'h' from equation (1)...

u'² = 2g × 2(u²/2g)

u'² = 4u²/2

u' = √2u²

u' = √2 u

Therefore, Option (C) is Correct.