Consider the motion of the cannonball along the vertical direction or y-direction.
= initial velocity along the Y-direction
= acceleration due to gravity
t = time of travel
Y = vertical displacement
Using the kinematics equation
Y =
t + (0.5)
t²
since the ball has been launched horizontally ,
= 0
Y = (0) t + (0.5)
t²
t =
![(2Y)/(a_(y))](https://img.qammunity.org/2020/formulas/physics/high-school/jmm44sl4uh6pkupuqlajroxyr3pw8p1gf5.png)
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
= initial velocity along the X-direction
= acceleration along the horizontal direction = 0
t = time of travel
X = horizontal displacement
Using the kinematics equation
X =
t + (0.5)
t²
since the ball has been launched horizontally ,
= 0
X =
t + (0.5) (0) t²
X =
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.