A body is projected horizontally from the top of a tower with initial velocity 18 ms -1- It hits the ground at angle 45- - What is the vertical component of velocity when it strikes the ground A- 9 ms -1B- 9 -2 m s-1C- 18 ms -1D- 18 -2 ms -1

The correct option is C When the body strikes the ground tan 45^∘=v_y/v_x=v_y/18=1 v_y=18 m/s ...

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Standard XII

Physics

Oblique Collision

A body is pro...

Question

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

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Solution

The correct option is C

When the body strikes the ground
$t a n 45^{\circ} = \frac{v_{y}}{v_{x}} = \frac{v_{y}}{18} = 1 v_{y} = 18 m / s$

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A body is projected horizontally from the top of a tower with initial velocity 18 ms−1. It hits the ground at angle 45∘. What is the vertical component of velocity when it strikes the ground

The correct option is C When the body strikes the ground tan45∘=vyvx=vy18=1vy=18 m/s

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

The correct option is C

When the body strikes the ground
$t a n 45^{\circ} = \frac{v_{y}}{v_{x}} = \frac{v_{y}}{18} = 1 v_{y} = 18 m / s$