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Compassionate
 one year ago
A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= 16t2 + 40ft + 1.5. Find the maximum height attained by the ball.
Compassionate
 one year ago
A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= 16t2 + 40ft + 1.5. Find the maximum height attained by the ball.

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Mehek14
 one year ago
Best ResponseYou've already chosen the best response.1\(\tt{x=\dfrac{b}{2a}}\)

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ b }{ 2a}\] \[\frac{ 40 }{ 32 }\] Time = 8

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0What is your equation exactly?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(h(t)= 16t^2 + 40t + 1.5\) ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0the max height will be the vertex (because the parabola opens down, and that is because the leading coefficient is negative)

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0I need to find the maximum height. Yes, that's the equation. So my time is 8, right? To find my height, plug 8 in.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0You don't need the height, if you are given that function you can find the vertex and that will be your maximum height.

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0To find max height, I need to first obtain the time it takes the ball to reach it, then plug in t.

Mehek14
 one year ago
Best ResponseYou've already chosen the best response.1how did you get time = 8? \(\tt{\dfrac{40}{32}=1.25}\) not 8

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0I got 8 because my calculator is autistic and I am jut realizing this. 16(1.25)^2 + 40(1.25) + 1.5 25 + 50 + 1.5 25 + 1.5 Height = 26.5

Mehek14
 one year ago
Best ResponseYou've already chosen the best response.1maximum height is 26.5 you get the time 1.25 and plug it in for t

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0Thanks! I got it :) `

Mehek14
 one year ago
Best ResponseYou've already chosen the best response.1when you use \(\tt{x=\dfrac{b}{2a}}\), you find the time then you plug the time in for the variable and yw :]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{black}{ \displaystyle h(t)=16t^240t+1.5 }\) \(\large\color{black}{ \displaystyle h(t)=16(t^2+2.5t)+1.5 }\) side caclulation: \(\large\color{black}{ \displaystyle 2.5\div2=1.25 }\) \(\large\color{black}{ \displaystyle 1.25^2=1.5625 }\) \(\large\color{black}{ \displaystyle h(t)=16(t^2+2.5t+1.56251.5625)+1.5 }\) \(\large\color{black}{ \displaystyle h(t)=16(t^2+2.5t+1.5625)+(16)(1.5625)+1.5 }\) \(\large\color{black}{ \displaystyle h(t)=16(t+1.25)^2+25+1.5 }\) \(\large\color{black}{ \displaystyle h(t)=16(t+1.25)^2+26.5 }\) So the height is 26.5 and it occurs at t=1.25.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0(via completing the square)

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0I never learned how to complete squares :(

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so can I shoot one more at you guys?

Mehek14
 one year ago
Best ResponseYou've already chosen the best response.1don't you learn completing the square in Alg 1?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0I don't think there is a single particular order for learning different concepts in math.... I have seen people who asked how to find a slope in DE section.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Well, completing the square is something that you can probably browse and find a good resource quite fast:)
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