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?
1 answer:
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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