Question

A particle is thorown downwards with a velocity of 20m/s from a height of 200 m high tower. at the same time another particle is thrown vertically find at what time and height above the ground these two particles​

Answer 1

Answer:

u=50m/s

g=−10m/s

2

The velocity at the highest point will be zero

Thus, v=0

Height can be calculated as,

v

2

−u

2

=2gs

0−2500=−20s

s=125m

Time can be calculated as,

v=u+gt

0=50+(−10)t

t=5sec