A 160.0kg projectile is fired off a 50.0m cliff with a velocity of 30.0m/s at 40.0 degree ATH.
a. Determine the time in the air before it hits the ground(in s)
b. Dtermine the distance from the bottom of the cliff it lands.(in m)
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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`
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Thank you for help! Could you please explain this part more specific for me?
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