Final answer:
The height of the lift when the ball meets it on its downward journey is 53 m. This is calculated using the initial velocity of the ball, the acceleration due to gravity, and the constant velocity of the lift.
Step-by-step explanation:
The question involves a scenario where a man is in a lift moving upward with a constant velocity and throws a ball upward relative to the lift. We want to determine the height of the lift when the ball meets it on its downward journey. To find this, we can use kinematic equations that describe motion with constant acceleration. The ball’s motion can be divided into two parts: its journey upward until it stops and its return journey back to the man’s hands. Since the lift is moving up with a constant velocity, this will not affect the time the ball takes to rise and fall relative to the man. Relative to the lift, the initial velocity of the ball is 5 m/s, and the acceleration due to gravity is -10 m/s2 (since it’s acting opposite to the ball’s motion).
Let's calculate the time the ball takes to rise and fall relative to the lift using the following kinematic equation for the upward journey where final velocity of the ball at the highest point is 0:
v = u + at
0 = 5 m/s - (10 m/s2 × t)
From this we get t (time to rise) = 0.5 s.
Now, the total time for the ball to return to the man's hand is double the time it takes to rise, which is 2 × 0.5 s = 1 s.
During this 1 second, the lift would have moved up an additional distance (since it is moving upwards with constant velocity):
d = velocity of the lift × time = 3 m/s × 1 s = 3 m.
Therefore, the height of the lift when it meets the ball on its downward journey is the initial height of the lift plus the additional distance it has moved up:
Height = initial height + additional distance = 50 m + 3 m = 53 m.