To solve this problem, we can use the equations of motion for vertical motion under gravity.
For the first ball: Initial velocity, u = 12 m/s (upward because it is thrown) Initial position, s = 24 m (above the ground) Acceleration due to gravity, g = 10 m/s² (downward)
Using the equation of motion for vertical displacement: s = ut + (1/2)gt²
Substituting the values: 24 = (12)t + (1/2)(10)(t²) 24 = 12t + 5t²
Rearranging the equation: 5t² + 12t - 24 = 0
To solve this quadratic equation, we can either factorize or use the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Substituting the values: t = (-(12) ± √((12)² - 4(5)(-24))) / (2(5)) t = (-12 ± √(144 + 480)) / 10 t = (-12 ± √624) / 10 t = (-12 ± 24.97) / 10
Using the positive value since time cannot be negative: t ≈ (12 + 24.97) / 10 ≈ 3.497 seconds
Therefore, the second ball must be dropped approximately 3.497 seconds later from the same starting point in order to hit the ground at the same time as the first ball.