Answer:
25.71 kgm/s
Explanation:
Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.
Given that K₂ = 0.7K₁
1/2mv₂² = 0.7(1/2mv₁²)
v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s
Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.
Th magnitude of object A's momentum change is thus 25.71 kgm/s
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
