Respuesta :
Answer:
a) # buses = 7 Â
Explanation:
For this exercise we use the kinematic equations, let's find the time it takes to reach the same height
 Â
   y =[tex]v_{oy}[/tex]  t - ½ g t²
Let's decompose the speed, with trigonometry
   v₀ₓ = v₀ cos θ
   [tex]v_{oy}[/tex] = v₀ sin  θ
   v₀ₓ = 40 cos 32
   v₀ₓ = 33.9 m / s
   [tex]v_{oy}[/tex] = 40 sin32
   [tex]v_{oy}[/tex] = 21.2 m / s
When it arrives it is at the same initial height y = 0
     0 = ([tex]v_{oy}[/tex] - ½ gt) t
That has two solutions
    t = 0           when it comes out
    t = 2 [tex]v_{oy}[/tex] / g    when it arrives
    t = 2 21.2 /9.8
    t = 4,326 s
We use the horizontal displacement equation
    x = vox t
    x = 33.9  4.326
    x = 146.7 m
To find the number of buses we can use a direct proportions rule
  # buses = 146.7 / 20
  # buses = 7.3
  # buses = 7
The distance of the seven buses is
   L = 20 * 7 = 140 m
b) let's look for the scope for this jump
   R = vo2 sin2T / g
   R = 40 2 without 2 32 /9.8
   R = 146.7 m
As we can see the range and distance needed to pass the seven (7) buses is different there is a margin of error of 6.7 m in favor of the jumper (security)