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chakrala mounica Grade: 11
        A bal is thrown horizontally at 9.8m/sec from a tower 40m high.The resultant velocity after 1 sec is....m/sec.
6 years ago

Answers : (2)

suryakanth AskiitiansExpert-IITB
105 Points
										

Dear chakrala,


In a horizontal projectile the horizontal velocity is constant and the vertical velocity changes because of the accelaration due to gravity


Hence Vx = 9.8 m/s


the verticla veloctiy after 1 sec is


Vy = -g*1      (the intial velocity in the verticla direction is zero)


Vy = -9.8 m/s


Hence the resultant velocity after 1 sec is


V = 9.8 m/s


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Suryakanth –IITB

6 years ago
vikas askiitian expert
510 Points
										

velocity in horizontal direction remains same


so Vx=9.8m/s which is constant


initially velocity in vertical direction is zero


velocity after 1 sec is Vy=Uy+at=0+9.8t            (a=g)


                                         Vy=9.8m/s


 V=9.8i-9.8j in vector form


   =9.8sqrt2 in magnitude

6 years ago
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