On earth, the height, h metres, of an object fired upward from the earth's surface at 48m/s is described by the equation h=48t-16t^2 where t seconds is the time since the object was fired upward. Determine: A.) the maximum height of the object B.) the times at which the object is 32m above the ground C.) the time at which the object hits the ground D.) the equation in vertex form PLEASE HELP ME :( I CAN'T ANSWER IT :(
Calculate how much tine it takes to reach the max height Vf=Vo-gt 0=48-(9.8)t 48=9.8t t=? Then replace t=?? In the function of height to find it
The time when it is 32m above is h=48t-16t^2 then h=32 32=48t-16t^2 we can divide all by 16 --->2=3t-t^2 solving
Why divide it to 16?
Dividing the equation by 16 is not required, but does reduce the problem a bit.
I feel like the max height is better don with differentiation
since you are asked for a max value of a function, take first direvative and set to 0 find the t and plug bakc in
Differentiation? That depends upon whether alfers101 has taken calculus or not. He may be in algebra. Please ask about current course work before making a statement like that.
I'm doing algebra
OK. To find the max height, find the vertex. Know the formula for the horiz. coordinate of the vertex? x=-b/(2a)
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