Quadratic inequality

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homeschool girl

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The height (in meters) of a shot cannonball follows a trajectory given by [MATH]h(t) = -4.9t^2 + 14t - 0.4[/MATH] at time [MATH]t[/MATH] (in seconds). As an improper fraction, for how long is the cannonball above a height of [MATH]6[/MATH] meters?

no clue where to start
 
Since h(t) is a negative quadratic (a=-4.9<0) then h(t) has a max point, ie it opens downwards. Please draw this and make sure your graph goes a good amount over 6.

Can you picture the t values we are looking for? Please draw any negative quadratic that goes above 6 and shade in the t-axis where this graph, call it h(t), has h(t) >6.

Please post back with this picture. Even if can't draw exactly what I am asking please post back some graph so we have a starting point.
 
i know how to graph it if its
[MATH]y = -4.9t^2 + 14t - 0.4[/MATH]but i don't understand how to solve the problem with the h(t)
 
Please post a graph and highlight where the h(t) is above 6ft. Highlight both where the graph is above 6ft and the t-values associated with those points. Please!
 
OK, thanks.
You really should label the t and h axes.

Where are the height, h, above 6ft. Highlight those points. Maybe even dray the line h=6. Everything above that has an h value more than 6.
 
homeschool girl said:
but there isn't an h axes, is there? theres only the vertical h*t axes and horizontal t axes
Click to expand...
Fair enough. There is an h(t) axis and a t axis. People also refer to the h(t) as the h axis.

Sorry about not explaining that, but thank you very much for telling me where I confused you.
 
Jomo said:
Fair enough. There is an h(t) axis and a t axis. People also refer to the h(t) as the h axis.

Sorry about not explaining that, but thank you very much for telling me where I confused you.
Click to expand...
oh so h(t) dosent mean h*t
 
ok.

when i try to solve it i get [MATH]h(t)=-4.9(t-\frac{10}{7})^2-9.6[/MATH]
wich woud mean that h dosent go above 0 so i guess i'm wrong
 
Last edited:
Lets suppose in my last graph, the line y = 6 and h(t) intersect at (a,6) and (b, 6).

Wouldn't the ball be above 6ft when t is between a and b, formally written as a<t<b?

If for example a= 5sec and b=12 seconds wouldn't the answer to your problem be 12s-5s = 7 seconds?

You just have to find a and b. That is you need to know when h(t) = 7 OR when h(t) - 7 =0. Can you solve h(t) - 7 for t?
 
homeschool girl said:
ok.

when i try to solve it i get [MATH]h(t)=-4.9(t-\frac{10}{7})^2-9.6[/MATH]
wich woud mean that h dosent go above 0 so i guess i'm wrong
Click to expand...
In your original definition of h(t) what is the value of h(0). In your equation above for h(t) what is h(0)? Are you getting the same result? You shouldn't be getting the same result!. Can we please your work so we can find your error? Also what is you ultimate goal in getting h(t) into that form?
 
Did you solve -4.9t^2+14t−0.4 - 6 = -4.9t^2 + 14t - 6.4 = 0?

What values did you get for t?
 
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