Asked by HELP
A projectile is thrown upward so that its distance above the ground after t sec is given by h(t)=-11t²+264t. After how many seconds does it reach its maximum height?
Answers
Answered by
Damon
11 t^2 - 264 t = - h
t^2 - (264/11 )t = -(1/11) h
t^2 - (24)t + 144 = -h/11 + 144
(t-12)^2 = -h/11 + 144
t = 12 at vertex
t^2 - (264/11 )t = -(1/11) h
t^2 - (24)t + 144 = -h/11 + 144
(t-12)^2 = -h/11 + 144
t = 12 at vertex
Answered by
Anonymous
A projectile is thrown upward so that its distance (in feet) above the ground after t seconds is given by
h left parenthesis t right parenthesis equals negative 15 t squared plus 420 t.
h(t)=−15t2+420t. What is its maximum height?
h left parenthesis t right parenthesis equals negative 15 t squared plus 420 t.
h(t)=−15t2+420t. What is its maximum height?
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