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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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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)

Posted by l70295100 on July 16, 2013 at 2:25 AM via web and tagged with physics, science

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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`

`30xx.643t-.5xx9.8xx t^2+50=0`

`19.29t-4.9t^2+50=0`

`t=(-19.29+-sqrt(19.29^2+50xx4xx4.9))/(-9.8)`

`t=(-19.29+-36.77)/(-9.8)`

`t=5.72` sec.

b. range of projectile (horizontal distance where it hit the ground)

`R=u cos(40^o)t`

`=30 xx cos(40^o)xx5.72=30xx.766xx5.72=131.45 m`

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Posted by aruv on July 16, 2013 at 3:18 AM (Answer #1)

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Thank you for help! Could you please explain this part more specific for me?

Thanks!

Posted by l70295100 on July 17, 2013 at 2:00 AM (Reply #1)