According to a certain estimate, the depth n(t), in centimeters, of the water in a certain tank at t hours past 2: 00 in the morning is given by n(t)= -20(t - 5)^2 + 500 for 0 ≤ t ≤ 10. according to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum? a) 5: 30 b) 7: 00 c) 7: 30 d) 8: 00 e) 9: 00

Ireally just need to get finished with precal! i’m really sorry i have to do this! hope this !

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward