Respuesta :
Answer:
a) t₁ = 70,717 s , b) x = 14143.4 m , c) θ = -60.02º
Explanation:
We will work on this problem using kinematics,
a) time to reach the ground, let's look for the initial speed
[tex]v_{oy}[/tex] = v₀ sin.60
v₀ₓ = v₀ cos 60
v_{oy} = 400 sin. 60 = 346.41 m / s
v₀ₓ = 400 cos 60 = 200 m / s
Let's use the projectile launch equation
y = y₀ + v_{oy} t - ½ g t²
Upon reaching the ground y = 0
0 = 1.5+ 346.41 t - ½ 9.8 t²
4.9 t² - 346.41 t - 1.5 = 0
t² - 70.696 t - 0.306 = 0
We solve the second degree equation
t = [70.696 ±√ (70.696² + 4 0.306)] / 2
t₁ = 70,717 s
t₂ = -0.0085 s
We take the positive time
b) let's use the equation
x = v₀ₓ t
x = 200 70,717
x = 14143.4 m
c) to find the angle let's look for the velocity at the point of impact, the velocity on the x-axis is constant
vₓ = v₀ₓ = 200 m / s
The speed on the y axis
[tex]v_{y}[/tex] = v_{oy} - g t
v_{y} = 346.41 - 9.8 70.717
v_{y} = -346.62 m / s
Let's use trigonometry
tan θ = v_{y} / vₓ
θ = tan⁻¹ v_{y} / vₓ
θ = tan⁻¹ (-346.62 / 200)
θ = -60.02º
