Question

the energy released in 2kg Wal is​

Answer 1

Initial k.e.=200 J

Let u be initial speed

(1/2)mu^2 =200 J

Given, m=2 kg

So, (1/2)×2u^2= 200

u^2= 200

At its highest point hmax, v=0

v^2 - u^2=- 2gh(minus because g is directed opposite to motion)

0–200=-2×g×hmax

Lets take g=10 m/secsec

So, hmax=200/(2×10)= 10m

Velocity half way up:

At halfway up ie h=5m

v^2 - u^2=- 2gh( minus because g is directed opposite to motion)

v^2 - 200=- 2×10×5=-100

v^2=100

v=10

Answer 2

I don’t know the answer





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