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ISRO Scientist ME 2010 Paper

- \(\frac{{\sqrt H }}{g}\)
- \(\sqrt {\frac{{2H}}{g}} \)
- \(\frac{{\sqrt {2H\sin \theta } }}{g}\)
- \(\frac{{\sqrt {2H} }}{{\sin \theta }}\)

Option 2 : \(\sqrt {\frac{{2H}}{g}} \)

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CT 10: Soil Mechanics

3808

10 Questions
10 Marks
10 Mins

__Concept:__

At the highest point of the projectile:

For projectile motion we know: a_{x }= 0 and a_{y }= -g

**Diagram**

**Explanation:**

V_{y }= 0 (at highest point)

\(V_y^2 - u_y^2 = - 2gH\)

u^{2 }sin ^{2}θ = 2gH

u sinθ = \(\sqrt {2{\rm{gH}}} \)

V_{y }= u_{y}+ a_{y}t

0 = \(\sqrt {2{\rm{gH}}} \) - gt

⇒ Time taken by the projectile to reach the highest point,

t = \(\sqrt {\frac{{2{\rm{H}}}}{{\rm{g}}}} \)

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