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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...

l70295100 | Student, Grade 12 | Honors

Posted July 16, 2013 at 2:25 AM via web

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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)

Tagged with physics, science

aruv | High School Teacher | Valedictorian

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

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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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l70295100 | Student , Grade 12 | Honors

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

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

Thanks!

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