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)

### 1 Answer | Add Yours

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

Thanks!

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