Question

13. If an object is thrown vertically up with the initial speed u from the ground,then the time taken by the object to return back to ground is

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Answer 1

Explanation:

Time of flight = 2u/g

Option D

Answer 2

» Case 1.

If body moves in upward direction then,

Final velocity (v) = 0

Acceleration (a) = - g [Because acceleration is in downward direction]

Initial velocity (u) = u

____________ [GIVEN]

• We have to find the time taken by the object to return back to the ground.

_______________________________

From first equation of motion

v = u + at

=> 0 = u + (-g) t

=> - u = - gt

=> u = gt

=> t = $\dfrac{u}{g}$

• For maximum height..

u² = 2gh

h = $\dfrac{ {u}^{2} }{2g}$ ______ (eq 1)

___________________________

Case 2.

If body is falling from maximum height then body moves downward.

Means..

Initial velocity (u) = 0

Final velocity (v) = v

Height/Distance (s) = - h

Acceleration (a) = - g

Time (t') = ?

We know that ..

s = ut + $\frac{1}{2}$ at²

=> - h = (0)t + $\frac{1}{2}$ (-g)(t')²

=> - h = $\frac{-1}{2}$ × g(t')²

=> $\dfrac{ {u}^{2} }{2g}$ = $\frac{1}{2}$ × g(t')²

=> (t')² = $\frac{ {u}^{2} }{ {g}^{2} }$

=> t' = $\dfrac{u}{g}$ _______ (eq 2)

______________________________

Now..

Total time (T) = t + t'

=> $\dfrac{u}{g}$ + $\dfrac{u}{g}$

=> $\dfrac{2u}{g}$

_______________________________

Option d) $\dfrac{2u}{g}$

_______ [ANSWER]

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