(a) The maximum height reached by the ball from the ground level is 75.87m
(b) The time taken for the ball to return to the elevator floor is 2.21 s
<u>The given parameters include:</u>
- constant velocity of the elevator, u₁ = 10 m/s
- initial velocity of the ball, u₂ = 20 m/s
- height of the boy above the elevator floor, h₁ = 2 m
- height of the elevator above the ground, h₂ = 28 m
To calculate:
(a) the maximum height of the projectile
total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)
at maximum height the final velocity of the projectile (ball), v = 0
Apply the following kinematic equation to determine the maximum height of the projectile.

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection
h = h₁ + h₂ + h₃
h = 28 m + 2 m + 45.87 m
h = 75.87 m
(b) The time taken for the ball to return to the elevator floor
Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m
Apply the following kinematic equation to determine the time to return to the elevator floor.

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In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
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