Answer:
Time = 0.86 s
Horizontal distance = 3.03 m
Explanation:
Given data:
initial velocity = [tex]v_{i}[/tex] = 3.8 m/s
θ = 22° below with horizontal
Height = h = 4.9 m
a.) Time = t = ?
b.) Horizontal distance = R = ?
a.) First we need to find time of flight
Resolve Velocity into horizontal and vertical component
Horizontal component = [tex]v_{i}_{x}[/tex] = [tex]v_{i}[/tex]Cosθ
                     = 3.8Cos22°
            [tex]v_{i}_{x}[/tex]   = 3.52 m/s
Vertical component = [tex]v_{i}_{y}[/tex]= [tex]v_{i}[/tex]Sinθ
                   = 3.8Sin22°
              [tex]v_{i}_{y}[/tex]   = 1.42 m/s
Using 2nd equation of motion
      [tex]h = v_{i}_{y}t+\frac{1}{2}gt^{2}[/tex]
         4.9 = 1.42t + 4.9t²
the above equation is quadratic. So it has 2 outputs. By solving above equation we have two outputs that is
         t = 0.86 s       &        t  = -1.15 s Â
Time can never be negative ,So the correct answer is t = 0.86 s
                 t = 0.86s
b.) Â Â
  As the horizontal component of velocity in projectile motion remain constant, so there is no acceleration along horizontal.
we can simply use this formula
         R = [tex]v_{i}_{x}t[/tex]
         R = (3.52)(0.86)
         R = 3.03 m    Â
