Answer:
1.03 seconds
Explanation:
x = x₀ + v₀ t + ½ at²
0 = 7.63 + (-2.40) t + ½ (-9.8) t²
0 = 7.63 - 2.40 t - 4.9 t²
Solve with quadratic formula:
t = [ -b ± √(b² - 4ac) ] / 2a
t = [ 2.40 ± √(2.40² - 4(-4.9)(7.63)) ] / -9.8
t = -1.52, 1.03
Since t can't be negative here, the sandbag hits the ground after 1.03 seconds.
