W24) A ball is dropped from rest from a height h above the ground. Another ball is thrown vertically
upwards from the ground at the instant the first ball is released. If the two balls are to meet at
a height of h/2 above the ground, then (a) how long does it take the two balls to meet at this
height, (b) what is the initial speed of the second ball when it meets the ball at this height and
(c) determine the speed of the second ball when they meet
a) Let's first consider the motion of the ball that was dropped from rest. The distance it has to travel is h/2 and it will be influenced by gravity. We can use the equation h = 0.5 * g * t^2, where h is the distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time it takes to reach that height.
h/2 = 0.5 * 9.8 * t^2
t = sqrt((h/2) / 4.9)
t = sqrt(h / 19.6)
Therefore, it takes sqrt(h / 19.6) seconds for the dropped ball to reach a height of h/2.
b) Now let's consider the motion of the ball thrown upwards. Since the two balls meet at the same height, the ball thrown upwards will also have to reach a height of h/2. We can use the kinematic equation v_f = v_i + at, where v_f is the final velocity, v_i is the initial velocity, a is the acceleration due to gravity, and t is the time taken.
For the ball thrown upwards:
h/2 = v_i * t - 0.5 * g * t^2
Since it reaches a height of h/2 when the dropped ball is also at h/2, we can substitute t = sqrt(h / 19.6) in the above equation:
h/2 = v_i * sqrt(h / 19.6) - 0.5 * 9.8 * (h / 19.6)
Solving for v_i gives:
v_i = sqrt(h * 9.8 - 9.8h) m/s
c) Now that we have the initial speed of the second ball, we can determine the speed of the second ball when they meet. Since the two balls meet at h/2 and the first ball is moving downward, we will consider the relative motion between the two balls.
At the point of meeting, the speed of the second ball will be the sum of their speeds. Therefore, the speed of the second ball when they meet will be equal to:
v_2 = v_i - 9.8 * sqrt(h / 19.6) + 9.8 * sqrt(h / 19.6)
v_2 = v_i
Therefore, the speed of the second ball when they meet will be the same as the initial speed of the second ball, which is sqrt(h * 9.8 - 9.8h) m/s.