Answer:
a) Θ = 18.5°
b) h = 6.26 m, 3.50 m, arrow goes over the branch
Step-by-step explanation:
having the following data:
Vo = 35 m/s
Θ = ?
horizontal distance = 75 m
a)
using the following equations:
Voy = Vo*(sin Θ) = 35*(sin Θ)
Vox = Vo*(cos Θ) = 35*(cos Θ)
horizontal distance to target = 75 = Vox*(2t); where t = Voy/g
replacing values:
75 = Vox*(2/g)*(Voy) = 2*(Vox)*(Voy)/9.8 =(Vo)²*[2*(sin Θ)*(cos Θ)]/g = (Vo)²(sin2Θ)/g
solving and using trigonometric identities:
sin2Θ = 75*(g)/(Vo)² = 75*(9.8)/(35)² = 0.6
2Θ = 36.91°
Θ = 18.5°
b)
The time to reach the maximum height will be equal to:
t = Voy/g = 35*(sin18.5°)/9.8 = 1.13 s
and the maximum height will be equal to:
h = 1/2gt² = (0.5)*(9.8)*(1.13)² = 6.26 m, 3.50 m, arrow goes over the branch