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A cannon is used by the department of forestry to launch containers of fire retardant chemical onto a fire. The cannon is a distance D=1 km from a cliff h=0.5 km high, and the fire is 1 km from the base of the cliff. If the retardant capsules are launched at an angle of 45, what initial speed v, is necessary?

Added Tue, 21 Jul '15

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Horizontal Component of Velocity = vCos45

= 0.707v

Horizontal Distance Covered , D = 1 Km

= 1000 m

Let it takes time t

Therefore

S = ut + 0.5*a*t^2

1000 = 0.707v*t + 0

t = 1414.21/v ----------------------(1)

The Verical Component of the Velocity = vSin45

= 0.707v

Therefore

Vertical Distance covered in time t = 0.5 - 1

= - 0.5 kM

= - 500 m

Therefore

S = ut + 0.5*g*t^2

-500 = 0.707vt - 0.5*9.8*t^2

Put the value of t from Eq(1) , we get

-500 = 0.707*(1414.21) - 0.5*9.8*(1414.21/v)^2

v = 80.833 m/sec

Added Tue, 21 Jul '15

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