Kinematics Question 644

Question: Three particles A, B and C are thrown from the top of a tower with the same speed. a is thrown up, b is thrown down and c is horizontally. They hit the ground with speeds vA , vB and vC respectively then,

Options:

A) vA=vB=vC

B)vA=vBvC

C) vAvCvB

D)vAvB=vC

Show Answer

Answer:

Correct Answer: A

Solution:

[a] For A: It goes up with velocity u will it reaches it’s maximum height (i.e. velocity becomes zero) and comes back to O and attains velocity u..

Using v2=u2+2asvA=u2+2gh

For B, going down with velocity u

vB=u2+2gh

For C, horizontal velocity remains same, i.e. u.

Vertical velocity =0+2gh=2gh

The resultant vC=vx2+vy2=u2+2gh.

Hence vA=vB=vC



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