If a ball is thrown off a 123m cliff at a speed of 18m/s ,how far from the base of the cliff is the ball ? show steps ill give brainliest

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

the answer would be the last one (constat) she stays at the same rat every two mintues it goes up by 0.4

step-by-step explanation: