1.The length of of a seconds hand in a watch is 1cm.The change in velocity of its tip in 15s is what?

2.A particle is projected vertically upwards and reaches the maxm. ht. H in time T.The ht. of a particle at any time t will be what?

Dear Chandan Das

speed of tip of second hand is constant =2∏*1/60  cm/sec

                                                         =∏/30  cm/sec

so in 15 sec it will rotate by an angel 90 degree

initial velocity = ∏/30 i

final velocity = ∏/30(-j)

change in velocity = | ∏/30(-j)  - ∏/30 i|

                          =∏√2/30  cm/sec

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Ans 2: Max Height: u^2sin^?/2g = H usin?/g = T Displacement at any time t = xtan?-(1/2)gx^2/(u^2cos^2?) Replace x with ucos?t : utsin?-(1/2)gt^2 =2Ht/T-(1/2)gt^2

Max Height Attained : H = (u²sin²θ)/2g      ½

 

Time Taken For Achieving Max. Height : T = usinθ/g

Displacement At Any time t: xtanθ-½gx²/(u²cos²θ)

Replace x with ucosθt : utsinθ-½gt²

= 2Ht/T-½gt²

 

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