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For part a) h(t)=-5t^2+7t =-5(t^2 - 7/5t) complete the square of the expression in parenthesis to put the function in the vertex form. Then comparing it to standard vertex form: h = a(t-h)^2 + k where (h,k) is the vertex. Identify k and that will be how high the ball goes.
I already told you I have .7
First you put that under b. Secondly, that is not k. You have h.
What do I do next?
plug in t = .7 in h(t)=-5t^2+7t and calculate h
there used to be constantly questions about slopes and lines now i see these questions a lot
when you throw a ball up in the air at the highest point its velocity = 0 as it is goes up going up to down, so somwhere in there it has 0 speed and velocity
so h'(t) = velocity set h'(t) = 0 and solve
that tells you the time it takes to get to highest veloicty then see what distance you travel given that time substituting that time into h(t)
that highest point*, with 0 velocity
oh is this precalc?
Try again. Don't forget there is a negative sign in front of t^2.
-5t^2+7t -5(0.7)^2 + 7(.7) = ?
Now you tell me...
You can evaluate that using your calculator.
okay, I got 2.45?
BTW, the notifications are not working for a couple of days and so people may not be notified when tagged or a reply is posted. Yes 2.45 meters is correct for a) For b) substitute h = 0.6 in h(t) = -5t^2+7t and solve for t. You will get two values for t -- the ball reaches 0.65 meters once while going up and once while coming down. Choose the higher t because they want the value when the ball is coming down.