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A shell is shot with an initial velocity v with arrow0 of 18 m/s, at an angle of θ0 = 60° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that the air drag is negligible?

Answer :

Answer:

D = 43 m

Explanation:

given,

initial velocity = 18 m/s

angle θ = 60°

total horizontal distance covered by the shell is

[tex]R = \dfrac{v_0^2sin 2\theta}{g}[/tex]

applying conservation of momentum in horizontal direction

m v₀ cos θ = m₁v₁ + m₂ v₂

m v₀ cos θ = 0.5 m v₂

v₂ = 2 v₀ cos θ.

distance covered by the shell from point of explosion

R' = v t

  = [tex](2 v_0 cos \theta) (\dfrac{v_0^2sin \theta}{g})[/tex]

  =[tex](2 \dfrac{v_0^2cos \theta sin \theta}{g})[/tex]

  = [tex]\dfrac{v_0^2sin 2\theta}{g}[/tex]

  = R

total distance traveled by the shell is

D = [tex]\dfrac{R}{2}+R'[/tex]

   = 1.5 R

   = [tex]1.5\dfrac{v_0^2sin 2\theta}{g}[/tex]

D = [tex]1.5\dfrac{18^2sin 2\times 60}{9.81}[/tex]

   = 42.9 ≅ 43 m

D = 43 m

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