this is my math problem

Question

Q#2 The heights (in feet) of a rocket T seconds after its launching is given by

                           H = -T³ + 96 T² +95 T + 20

Find out

 i). The Speed or velocity of the rocket at any time T.

ii). The Speed or velocity of the rocket when T= 0, 40, 80 seconds and interpret the results.

iii). The maximum height attained by the Rocket.                                                     

Answer 1

i) v(T) = H'(T) = - 3T² + 192T + 95; this is velocity and speed Iv(T)I = I- 3T² + 192T + 95I

ii) v(0) = 95 ft/s; v(40) = 2975 ft/s; v(80) = - 3245 ft/s

Iv(0)I = 95 ft/s; Iv(40)I = 2975 ft/s; Iv(80)I = 3245 ft/s

iii) v(T) = 0; 3T² - 192T - 95 = 0; T = - 0.491 or T = 64.491 s

max H(T) = H(64.491) = 137195.38 ft