A cracker is thrown into air with a velocity of 10 m/s at an angle of 450with the vertical. When it is at a height of 0.5 m from the ground, it explodes into a number of pieces which follow different parabolic paths. What is the velocity of centre of mass, when it is at a height of 1 m from the ground? (g = 10 m/s2)
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the center of mass follow thw same path and the velocity as it was not exploded .u=10m/s ux=10cos45=102√ m/suy=102√ m/svy=1002−2*10‾‾‾‾‾‾‾‾‾‾‾‾√=30 ‾‾‾√ m/svelocity of CM at height 1m v=1002+30‾‾‾‾‾‾‾‾√=80‾‾‾√ m/s
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