Consider the motion of the cannonball along the vertical direction or y-direction.
[tex]v_{oy}[/tex] = initial velocity along the Y-direction
[tex]a_{y}[/tex] = acceleration due to gravity
t = time of travel
Y = vertical displacement
Using the kinematics equation
Y = [tex]v_{oy}[/tex] t + (0.5) [tex]a_{y}[/tex] t²
since the ball has been launched horizontally , [tex]v_{oy}[/tex] = 0
Y = (0) t + (0.5) [tex]a_{y}[/tex] t²
t =[tex]\frac{2Y}{a_{y}}[/tex]
hence the time of travel is independent of the initial velocity. hence the time of travel remain same.
Consider the motion along the horizontal direction
[tex]v_{ox}[/tex] = initial velocity along the X-direction
[tex]a_{x}[/tex] = acceleration along the horizontal direction = 0
t = time of travel
X = horizontal displacement
Using the kinematics equation
X = [tex]v_{ox}[/tex] t + (0.5) [tex]a_{x}[/tex] t²
since the ball has been launched horizontally , [tex]a_{x}[/tex] = 0
X = [tex]v_{ox}[/tex] t + (0.5) (0) t²
X = [tex]v_{ox}[/tex] t
hence the horizontal distance directly depends on the velocity. so the horizontal distance will become double.
C) The time won't change, but horizontal distance will double.
Answer:
C) The time won't change, but horizontal distance will double.
Explanation:
The time won't change, but horizontal distance will double. The time is based on the acceleration due to gravity (g) and the vertical distance, so it will not change. If the horizontal velocity changes, the horizontal distance will double based on the formula d=tv.
