Prove that the trajectory of a projectile is parabolic, having the form y = ax + bx2. to obtain this expression, solve the equation x = v0xt for t and substitute it into the expression for y = v0yt − 1 2 gt2. (these equations describe the x and y positions of a projectile that starts at the origin.) you should obtain an equation of the form y = ax + bx2 where a and b are constants.

y = v₀tgθx - gx²/2v₀²cos²θ

a = v₀tgθ

b = -g/2v₀²cos²θ

Step-by-step explanation:

x = v₀ₓt

y = v₀y.t - g.t²/2

x = v₀.cosθt → t = x/v₀.cosθ

y = v₀y.t - g.t²/2

v₀y = v₀.senθ

y = v₀senθ.x/v₀cosθ - g/2.(x/v₀cosθ)²

y = v₀.tgθ.x - gx²/2v₀²cos²θ

a = v₀tgθ → constant because v₀ and θ do not change

b = - g/2v₀²cos²θ → constant because v₀, g and θ do not change

answer: you have to send a photo of the question

step-by-step explanation:

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

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