Two boys simultaneously aim their guns at a bird sitting on a tower. The first boy releases his shot with a speed of 100km/s at an angle of projection of 30. The second boy is ahead of the first boy by a distance of 50m and releases his shot with a speed of 80m/s. how must he aim his gun so that both the shots hit the bird simultaneously? What is the distance of the foot of the tower from the two boys and the height of the tower? With what velocities and when do the two shots hit the bird?

Answer 1

Suppose that the bullet hits the aim at the upward motion, then
h= v₀₁•sinα•t - gt²/2 =
v₀₂•sinβ•t - gt²/2,
v₀₁•sinα = v₀₂•sinβ ,
sinβ= v₀₁•sinα/ v₀₂=
=100•sin30⁰/80 =0.625,
β=sin⁻¹0.625 = 38.7⁰.

x₁-x₂=Δx,
v₀₁•cosα•t = v₀₂•cosβ•t = Δx,
t= Δx/{ v₀₁•cosα - v₀₂•cosβ}=
=50/{100•cos30⁰ -80•cos38.7⁰}=
=2.07 s.

h=v₀₁•sinα•t - gt²/2=
=100•0.5•2.07 = 9.8•2.07²/2 =82.5 m.

x₁=v₀₁•cosα•t =100•cos30⁰•2.07 =179.3 m
x₂=x₁-Δx = 179.3-50 =129.3 m.

v₁(y)=v₀₁•sinα - g•t =
=100•0.5 – 9.8•2.07 =29.7 m/s.
v₁(x)= v₀₁•cosα =100•0.866 =86.6 m/s,
v₁= sqrt{v₁(x)²+v₁(y)²} =
=sqrt{29.7²+86.6²} = 91.6 m/s.

v₂(y)=v₀₂•sinβ - g•t =
=80•0.625 – 9.8•2.07 =29.7 m/s.
v₂ (x)= v₀₂•cosβ =80•0.78 =62.4 m/s,
v₂= sqrt{v₂ (x)²+v₂ (y)²} =
=sqrt{29.7²+62.4²} = 69.1 m/s.