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Bharadwaj Vemparala Grade: 11
        

A ball is thrown from ground into air. At a height 9.1m, the velocity is observed to be V=7.6i+6.1j.Then find:


(a)Maximum height,


(b)Horizontal range and


(c)velocity of the ball at the instant of striking the ground.

7 years ago

Answers : (1)

Anurag Kishore
37 Points
										

Hi


As per question ux = 7.6 m/s     uy = 6.1 m/s


Horizontal component will remain the same throughout the motion because there is no force in x direction


(a) Use the eqn


      vy2 = uy2 + 2 ay s


=> 0 = (6.1)2 - 2* 9.8 s


=> s = 1.9m


So,


Maximum height = 9.1 + 1.9 = 11 m


 


(b) Use


      sy = uyt + 1/2 ay t2


=> 11 = 0 + 1/2 * 9.8 t2             (at heighest point uy = 0)


=> t = 1.5 sec


=> dx = uxt = 7.6 * 1.5 = 11.4 m


 


(c) At the instant of striking


   vx = ux = 7.6 m/s


   vy =ay t = 9.8 * 1.5 = 14.7 m/s


 


 


Thanks


Anurag Kishore

7 years ago
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