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^@`. What is the vertical component of velocity when it strikes the ground ?

Options

A. `9 m s^-1`
B. `9 sqrt(2) m s^-1`
C. `18 m s^-1`
D. `18 sqrt(2) m s^-1`

Answer

C

Explanation

( c) `v cos 45^@ = u = 18 ms^-1`
`rArr v = 18 sqrt(2) ms^-1`
Vertical component
`v sin 45^@ = 18 sqrt(2) xx (1)/(sqrt(2) = 18 ms^-1`.
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