A body is projected horizontally from top of a tower with initial velocity 18,m s^{-1} . It hits gro

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3-2.Motion in Plane

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A body is projected horizontally from the top of a tower with initial velocity $18\,m s^{-1}$. It hits the ground at angle $45^o$. What is the vertical component of velocity when it strikes the ground ......... $ms^{-1}$

A

$9$

B

$9\sqrt 2$

C

$18$

D

$18\sqrt 2$

Solution

When the body strikes the ground,

$\tan 45^{\circ}=\frac{v_{y}^{\prime}}{v_{x}} \Rightarrow \frac{v_{y}}{18}=1$

$\Rightarrow v_{y}=18 m s^{-1}$

Standard 11

Physics

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