Question

A body is projected from a tower of height 500m horizontally with a speed 20m/s The approximate displacement of the body after 5s is

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Answer 1

Answer:

Since, body is projected horizontally with a velocity of 20m/s.

ux=20m/s

ax=0

So, vx=20m/s for all time.

Along vertical axis,

uy=0

t=5

$$ay = 10m/s2$

vy=?

vy=uy+ayt

=>vy=0+10∗5=50m/s

So, magnitude of velocity of body after $$5s$

{(50)

2

+(20)

2

}

m/s

=

2900

m/s

=53.851m/s

=54m/s.

Answer 2

Answer:

80 √2m

Explanation:

υ=20 m/s

t=4 sec

s

h

=v

h

×t=20×4=80m

υ

v

=0

g=10 m/s

2

t=4 sec

s

v

=

2

1

×10×4×4

s

v

=80m

Displacement, d=

s

v

2

+s

n

2

=80

2

m

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