Respuesta :
Answer:
a) y = 127.6 m, b) 11.9s, c) v = 103.6 m / s, θ’= 326.7º
Explanation:
This is a missile throwing exercise
let's start by breaking down the initial velocity
sin 30 = [tex]\frac{v_{oy} }{v_o}[/tex]
cos 30 = v₀ₓ / v₀
v_{oy} = v₀ go sin 30
v₀ₓ = v₀ cos 30
v_{oy} = 100 sin 30 = 50 m / s
v₀ₓ = 100 cos 30 = 86.6 m / s
a) the maximum height is requested.
At this point the vertical velocity is zero (v_y = 0)
v_y² = [tex]v_{oy}^2[/tex] - 2 g y
0 = v_{oy}^2 - 2g y
y = [tex]\frac{v_{oy}^2 }{2g}[/tex]
y = 50² / (2 9.8)
y = 127.6 m
b) Flight time
this is the time it takes to reach the ground, the reference system for this movement is taken on the ground this is a height of y = 0 m and the body is at an initial height of i = 100m
y = y₀ + v₀ t - ½ g t²
0 = 100 + 50 t - ½ 9.8 t²
we solve the quadratic equation
4.9 t² - 50 t - 100 = 0
t = [tex]\frac{50 \pm \sqrt{50^2 + 4 \ 4.9 \ 100} }{2 \ 4.9}[/tex]
t = [tex]\frac{50 \ \pm \ 66.8}{9.8}[/tex]
t₁ = 11.9 s
t₂ = -8.4 s
flight time is 11.9s
c) The time 1 s before hitting the ground is
t1 = 11.9 -1
t1 = 10.9 s
let's find the vertical speed
v_y =[tex]v_{oy}[/tex] - g t
v_y = 50 - 9.8 10.9
v_y = -56.8 m / s
the negative sign indicates that the direction of the velocity is downward.
On the x-axis there is no acceleration therefore the speed is constant.
Let's use the Pythagorean theorem
v = [tex]\sqrt{v_x^2+v_y^2}[/tex]
v = [tex]\sqrt { 86.6^2 + 56.8^2}[/tex]
v = 103.6 m / s
let's use trigonometry
tan θ = [tex]\frac{v_y}{v_x}[/tex]
θ = tan⁻¹ \frac{v_y}{v_x}
θ = tan⁻¹ (-56.8 / 86.60)
θ = -33.3º
the negative sign indicates that it is measured clockwise from the x-axis
for a counterclockwise measurement
θ’= 360 - θ
θ' = 360 - 33.3
θ’= 326.7º
