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A 160.0kg projectile is fired off a 50.0m cliff with a velocity of 30.0m/s at 40.0...
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height of cliff=50.0 m
u=initial velocity= 30.0 m/s
angle of projection= 40 degree
a. Time of flight
vertical component of velocity is usin(40degree)
Let t be the time of flight .e before it reaches ground ,its velocity will be zero.
`u sin(40^o)t-(1/2)g t^2+50=0`
b. range of projectile (horizontal distance where it hit the ground)
`=30 xx cos(40^o)xx5.72=30xx.766xx5.72=131.45 m`
Posted by aruv on July 16, 2013 at 3:18 AM (Answer #1)
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