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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?​

Respuesta :

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

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