consider the motion of the ball along the vertical direction
v₀ = initial velocity in vertical direction = 0 m/s (since the ball was thrown horizontally)
a = acceleration = acceleration due to gravity = 9.8 m/s²
Y = vertical displacement = height of the cliff = 123 m
t = time of travel
Using the kinematics equation
Y = v₀ t + (0.5) a t²
inserting the values
135 = (0) t + (0.5) (9.8) t²
t = 5.25 sec
consider the motion of the ball along the horizontal direction
v'₀ = initial velocity in horizontal direction = 18 m/s
a' = acceleration = 0 m/s²
X = horizontal displacement = ?
t = time of travel = 5.25 sec
Using the kinematics equation
X = v'₀ t + (0.5) a' t²
inserting the values
X = (18) (5.25) + (0.5) (0) t²
X = 94.5 m
