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A bal is thrown horizontally at 9.8m/sec from a tower 40m high.The resultant velocity after 1 sec is....m/sec.

A bal is thrown horizontally at 9.8m/sec from a tower 40m high.The resultant velocity after 1 sec is....m/sec.

Grade:11

2 Answers

suryakanth AskiitiansExpert-IITB
105 Points
10 years ago

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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Askiitians Expert

Suryakanth –IITB

vikas askiitian expert
509 Points
10 years ago

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

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