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mell65
Group Title
A ball is kicked upward with an initial velocity of 56 feet per second. The ball's height h (in feet) can be expressed as a function of time t (in seconds) by the equation h = 16t2 + 56t. How much time does the ball take to reach its highest point?
 2 years ago
 2 years ago
mell65 Group Title
A ball is kicked upward with an initial velocity of 56 feet per second. The ball's height h (in feet) can be expressed as a function of time t (in seconds) by the equation h = 16t2 + 56t. How much time does the ball take to reach its highest point?
 2 years ago
 2 years ago

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mark_o. Group TitleBest ResponseYou've already chosen the best response.0
@ mell6 are in calculus class? to get to the max height, differentiate h, that will be you velocity, v=dh/dt now set this to zero, dh/dt=0, then solve for t, then sub this t to h, that is the max h
 2 years ago

mark_o. Group TitleBest ResponseYou've already chosen the best response.0
are u in calculus class?
 2 years ago

mell65 Group TitleBest ResponseYou've already chosen the best response.0
no but i got it it's like 1.25 i think
 2 years ago

mark_o. Group TitleBest ResponseYou've already chosen the best response.0
t =1.75 sec sub this to h to get hmm max h= 49 ft? yes or no?
 2 years ago

mark_o. Group TitleBest ResponseYou've already chosen the best response.0
so you are not in calculus class? h = 16t^2 + 56t. we noticed that this is in the family of parabola y=ax^2 +b or y=16x^2 +56x plot this on the graph the parabola is opening downward and the vertex is V=b/2a we get V=56/2(16)=1.75 =x=t in sec now sub this to h = 16t2 + 56t we get max h=?
 2 years ago

mark_o. Group TitleBest ResponseYou've already chosen the best response.0
@ mell65 let me know if you have question
 2 years ago
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