A pellet was shot vertically upward with a velocity of \(25 \frac{\text{m}}{\text{s}}\) from a height of \(4 \text{ m}\). If the acceleration of gravity is \(5 \frac{\text{m}}{\text{s}^2}\), at what time will the pellet reach the height of \(9\text{ m}\) going downward.

[Use the formula \(h_{(t)} = h_0 + V_0t-at^2\), where \(h_0\) = initial height, \(v_0\) = initial velocity, \(a\) = acceleration, and \(t\) = time]