consider the motion of projectile A in vertical direction :
v₀ = initial velocity of projectile A in vertical direction = 0 m/s (since the projectile was launched horizontally)
a = acceleration of the projectile = g = acceleration due to gravity = 9.8 m/s²
t = time of travel for projectile A = 3.0 seconds
Y = vertical displacement of projectile A = height of the cliff = h = ?
using the kinematics equation along the vertical direction as
Y = v₀ t + (0.5) a t²
h = (0) (3.0) + (0.5) (9.8) (3.0)²
h = 44.1 m
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