Question

A body is projected horizontally from the top of a tower with a velocity of 10m/s if it hits the ground at an angle of 45° then the verticle component of velocity when it hits ground is

Answer 1

Answer:

first thing u should know is that whenever u r doing projectile motion u have the vertical component as ucos theta.

As per the question we have,

u= 10 m/s

theta = 45 degree so

ucos theta = 10 * cos 45 degree

                  = 10 * 1 upon root 2

                  =  10÷√2

                  =   8.33

Explanation:

Answer 2

Answer:

Let the body hit the ground with velocity v and its vertical componant of velocity be v

As no external force acts on the body in horizontal direction, thus the horizontal velocity of the body remains the same i.e

=10 m/s

∴ vsin45

o

=v

Also vcos45

o

=10

Dividing (1) and (2) we get,

10

v

=tan45

o

=1

⟹v

=10 m/s

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