Respuesta :
Answer:
d) None of the above
Explanation:
[tex]v_{o}[/tex] = inituial velocity of launch = 4 m/s
θ = angle of launch = 10 deg
Consider the motion along the vertical direction
[tex]v_{oy}[/tex] = initial velocity along vertical direction = 4 Sin10 = 0.695
m/s
[tex]a_{y}[/tex] = acceleration along the vertical direction = - 9.8 m/s²
y = vertical displacement = - 20 m
t = time of travel
using the equation
[tex]y=v_{oy} t+(0.5)a_{y} t^{2}[/tex]
- 20 = (0.695) t + (0.5) (- 9.8) t²
t = 2.1 sec
consider the motion along the horizontal direction
x = horizontal displacement
[tex]v_{ox}[/tex] = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s
[tex]a_{x}[/tex] = acceleration along the horizontal direction = 0 m/s²
t = time of travel = 2.1 s
Using the kinematics equation
[tex]x =v_{ox} t+(0.5)a_{x} t^{2}[/tex]
x = (3.94) (2.1) + (0.5) (0) (2.1)²
x = 8.3 m
Consider the motion along the vertical direction
[tex]v_{oy}[/tex] = initial velocity along vertical direction = 4 Sin10 = 0.695
m/s
[tex]a_{y}[/tex] = acceleration along the vertical direction = - 9.8 m/s²
[tex]y_{o}[/tex] =initial vertical position at the time of launch = 20 m
[tex]y[/tex] = vertical position at the maximum height = 20 m
[tex]v_{fy}[/tex] = final velocity along vertical direction at highest point = 0 m/s
using the equation
[tex]{v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})[/tex]
[tex]0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)[/tex]
[tex]y[/tex] = 20.02 m
h = height above the starting height
h = [tex]y[/tex] - [tex]y_{o}[/tex]
h = 20.02 - 20
h = 0.02 m
