A water balloon is fired at 34.0 m/s from a water cannon, which is aimed at an angle of 18° above the ground. The centre of the cannon’s target (which has a radius of 1.0 m) is painted on the asphalt 42.0 m away from the water cannon. Will the balloon hit the target? Justify your response with calculations that indicate where the water balloon will land and make one suggestion about how to adjust the water cannon so that the water balloon will hit the target.
Grade 11 Physics: I haven't yet learned the range formula but don't know how to adjust the angle to hit the target by using the formulas I have used.
A water balloon is fired at 34.0 m/s from a water cannon, which is aimed at an angle of 18° above the ground. The centre of the cannon’s target (which has a radius of 1.0 m) is painted on the asphalt 42.0 m away from the water cannon. Will the balloon hit the target? Justify your response with calculations that indicate where the water balloon will land and make one suggestion about how to adjust the water cannon so that the water balloon will hit the target.
sin??θ=opphyp
sin??θ=v⃗yv⃗
v⃗y=v⃗sinθ
=(34.0m/s)sin 18°
=10.5m/s[up]
cosθ=adjhyp
cosθ=v⃗xv⃗
v⃗x=v⃗cosθ
=(34.0m/s)cos 18°
=32.3m/s[forward]
Let [up] be positive
Δd⃗=0
a⃗=-9.8m/s2[down]
v⃗y1=10.5m/s[up]
Δt= ?
Δd⃗=v⃗y1Δt+12a⃗(Δt)2
0=(10.5m/s)Δt+12(-9.8m/s2)(Δt)2
0=(10.5m/s)Δt-4.9m/s2(Δt)2
0=Δt (10.5m/s-4.9m/s2Δt)
Δt=0 or Δt=2.14s
v⃗x=32.3m/s [forward]
Δt=2.14s
Δd⃗x= ?
Δd⃗x=v⃗xΔt
=(32.3m/s)(2.14s)
=69m[forward]
The ball will land after 69m and will miss the target by 27m.
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