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A bomb at rest is exploded and the pieces are scattered in all directions with a maximum velocity of 20m/s dangerous distance from that spot is (g=10)

Thousif MT , 9 Years ago
Grade 12th pass
anser 3 Answers
Vikas TU
The dangerous distance from the spot would be:
From the Newton’s first law of equation:
20^2 = 0^2 + 2gs
s = 400/2g => 20 meter.
Last Activity: 9 Years ago
??? ????
max range = (u^2 * sin2x)/g
so for x=45
dangerous distance = 400/10 = 40 m 
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Last Activity: 9 Years ago
Balaji
Particles may scatter in any direction. Horizontal range = (u*u*sin2a)/g where a is angle made with horizontal. It`s maximum value will be when a=45 degree. Substituting the values, maximum range will be 40m. Distances up to 40m are dangerous.
Last Activity: 9 Years ago
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