a projectile is thrown upward so that it's distance above the ground after T seconds is H equals - 16t^2 + 672 T. After how many seconds does it reach its maximum height?

Step-by-step explanation:

T = time in seconds

H = distance

Tground = time to return to ground

Tmax = time at maximum height

H = - 16T^2 + 672T ... eqn 1

projectile returns to ground at H = 0,

subs for H in eqn 1 ...

0 = - 16T^2 + 672T

solving for T we get ...

16T^2 = 672T

=> Tground = 42secs

Tmax = 0.5 Tground = 21secs