Question

a particle is projected vertically upward from the top of a tower of height 100m with an initial speed of 100m/s. the distance travelled by the particle during 21st second

Answer 1

Answer:

Relative displacement =100 m

Relative acceleration =0 m/s^2

 because both have acceleration g

Relative initial velocity =−50 m/s

Using relative motion,

s=ut+(1/2)at^  2

100=50t+(1/2)×0×t^  2  

t=2 s

For particle projected from ground,

s=ut+(1/2)at^  2

s=50×2−(1/2)×10×2^  2  =80 m

And this is a PHYSICS question not of MATHS next time be careful......

Hope the answer is clear and mark it as brainliest pls

Step-by-step explanation:

Answer 2

Answer:

hiiii

Step-by-step explanation:

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