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
Time can never be negative ,So the correct answer is t = 0.86 s
t = 0.86s
b.)
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
