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Grade 12Mechanics

three particles A ,b and C are thrown from the top of a tower with the same speed .A is thrown straight up.B isn thrown straight down and C is thrown horizontly they hit the ground with speed v_{A ,}V_B,and v_cthen compare the three velocities

abhay tiwari

9 Years agoGrade 12

3 Answers

SAHIL

Approved Tutor Answer9 Years ago

When Particle A arrives back at the top of the tower, it will have the same downward speed as the upward speed it started with; its subsequent motion will be identical to that of Particle B. That va and vb are > vc is clear from the fact that A is freely "dropped" from its maximum height, which is greater than the height of the tower, hence greater than max height of C.

SO ANSWER IS

va = vb > vc

THANKS

Guru

9 Years ago

For A , Va = u + gt

For B, Vb = u + gt

For c, Vx = u ( g= 0 )

Vy = gt ( Uy = 0 )

For Vc = root Vx^2 + Vy^2

= root u^2 + gt ^2

= u + gt

Hence Va = Vb = Vc.

Rajat

8 Years ago

Va^2 =U^2 +2(-g)(-h)=U^2 +2ghVb^2 =U^2 +2ghFor CVc in x direction=UVc in y direction=√2ghVc^2= (Vc)x^2 + (Vc)y^2 =U^2 + 2ghVa^2=Vb^2= Vc^2Va=Vb= Vc

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