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A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (a) Determine the time the projectile takes to hit the ground. (b) Find the distance from the base of the cliff to the point where projectile hits the ground. (c) Just before the projectile hits the ground: Find i) the horizontal and vertical components of its velocity, ii) magnitude and direction of the velocity vector i.e. angle it makes with the horizontal. (d) Find the maximum height above the cliff top reached by the projectile.

Respuesta :

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants ([tex]v_{oy}[/tex] = 0) zero, so let's use the equation

     y =[tex]v_{oy}[/tex] t -1/2 g t²

     y= - ½ g t²

     t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

     x = vox t

     x = v₀ₓ √2y/g

c) Speeds before touching the ground

     vₓ = vox = constant

     [tex]v_{y}[/tex] = [tex]v_{oy}[/tex] - gt

     [tex]v_{y}[/tex] = 0 - g √2y/g

    [tex]v_{y}[/tex]  = - √2gy

    tan θ = Vy / vx

    θ = tan⁻¹ (vy / vx)

    θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

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