when serving in tennis a player tosses the tennis ball vertically in the . The height h of the ball after t seconds is given by the quadratic function h(t)=-5t^2+7t (the height is measured in meters from the point of the toss.)
a. How high in the air does the ball go?
b. Assume that the player hits the ball on its way down when it's 0.6m above the point of the toss. For how many seconds is the ball in the air between the toss and the serve?
I've already got -7/2x(-5)=7/10-.7
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For part a)
=-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.
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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
@ranga I got 5.145 for h. Is that right?
oh is this precalc?
Try again. Don't forget there is a negative sign in front of t^2.
-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.