To solve this problem, we can use the kinematic equations of motion. Let's analyze the motion of each ball separately.
For the first ball thrown upward: Initial velocity (u) = 12 m/s (upward) Final velocity (v) = 0 m/s (at the highest point) Acceleration (a) = -9.8 m/s² (acceleration due to gravity, pointing downward) Initial position (s) = 24 m (above the ground)
We can use the kinematic equation: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Substituting the values: 0^2 = (12)^2 + 2(-9.8)s 0 = 144 - 19.6s 19.6s = 144 s = 144 / 19.6 s ≈ 7.35 m
The first ball will reach a maximum height of approximately 7.35 meters above the starting point.
Now, for the second ball dropped from the same starting point: Since the ball is dropped, its initial velocity (u) is 0 m/s. Acceleration (a) is still -9.8 m/s² (acceleration due to gravity, pointing downward). We need to find the time it takes for the second ball to reach the ground.
We can use the kinematic equation: s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
Substituting the values: 24 = 0 * t + (1/2) * (-9.8) * t^2 24 = -4.9t^2 t^2 = -24 / -4.9 t^2 ≈ 4.898 t ≈ √4.898 t ≈ 2.21 s
Therefore, the second ball must be dropped approximately 2.21 seconds later to hit the ground at the same time as the first ball.