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# 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.

11 years ago

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