Question

Suppose there is an object released from a height of 500m. What is the velocity of object after first one second?

Answer 1

Hey.

Object is released from height of 500 m

so, initial velocity of the object = 0 m/s

a = g = 10 m/s^2

t = 1 sec

As s = u t + 1/2 a t^2

so, s = 0 + 1/2 ×10×1^2 = 5 m from the top

so, it's still in the air

As v = u + a t

So, velocity after 1 sec

v = 0 + 10 × 1 = 10 m/s

Thanks.

Answer 2

Heya....

As object is released from ...

Hight = 500m
Initial velocity = 0 m/s
a = g = 10m/s^2
time taken = 1 s

Now,, distance is...

s = ut + 1/2 at^2

= 0+1/2x 10x1^2
= 5 m...

Still in air,,

V = u + at
0+10x1 = 10 m/s....

-- Be Brainly...