Respuesta :
Answer:
the shooting angle ia 18.4º
Explanation:
For resolution of this exercise we use projectile launch expressions, let's see the scope
R = Vo² sin (2θ) / g
sin 2θ = g R / Vo²
sin 2θ = 9.8 75/35²
2θ = sin⁻¹ (0.6)
θ = 18.4º
To know how for the arrow the tree branch we calculate the height of the arrow at this point
X2 = 75/2 = 37.5 m
We calculate the time to reach this point since the speed is constant on the X axis
X = Vox t
t2 = X2 / Vox = X2 / (Vo cosθ)
t2 = 37.5 / (35 cos 18.4)
t2 = 1.13 s
With this time we calculate the height at this point
Y = Voy t - ½ g t²
Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²
Y = 6.23 m
With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch
Answer:
a) Θ = 18.5°
b) h = 6.26 m, 3.50 m, arrow goes over the branch
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
