help

- anonymous

help

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- anonymous

@xapproachesinfinity

- anonymous

@Michele_Laino

- anonymous

whats your question?

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## More answers

- anonymous

The function H(t) = -16t2 + vt + s shows the height H (t), in feet, of a projectile launched vertically from s feet above the ground after t seconds. The initial speed of the projectile is v feet per second.
Part A: The projectile was launched from a height of 100 feet with an initial velocity of 60 feet per second. Create an equation to find the time taken by the projectile to fall on the ground. (2 points)
Part B: What is the maximum height that the projectile will reach? Show your work. (2 points)
Part C: Another object moves in the air along the path of g(t) = 20 + 38.7t where g(t) is the height, in feet, of the object from the ground at time t seconds.
Use a table to find the approximate solution to the equation H(t) = g(t), and explain what the solution represents in the context of the problem? [Use the function H(t) obtained in Part A, and estimate using integer values] (4 points)
Part D: Do H(t) and g(t) intersect when the projectile is going up or down, and how do you know? (2 points)

- anonymous

i wish i could help with this one but im not sure sorry :(

- anonymous

thanks for trying

- Michele_Laino

using your initial conditions, we can rewrite your equation, like below:
\[\Large H\left( t \right) = - 16{t^2} + 60t + 100\]

- Michele_Laino

furthermore the equation for speed v(t) is:
\[\Large v\left( t \right) = 60 - 32t\]

- Michele_Laino

when our projectile is going up, it will reach the maximum height, and at that time its speed is zero, so we can write this:
\[\Large 0 = 60 - 32{t_1}\]
where t_1 is the time needed to projectile in order to reach the maximum height

- Michele_Laino

solving that equation for t_1 we have:
\[\Large {t_1} = \frac{{60}}{{32}} = ...\sec \]

- Michele_Laino

please complete that computation

- Michele_Laino

please wait...

- anonymous

no problem you are a lifesaver

- Michele_Laino

ok! I'm here

- Michele_Laino

we get:
\[\Large {t_1} = \frac{{60}}{{32}} = 1.875\sec \] am I right?

- anonymous

sure

- Michele_Laino

now the maximum height reached by our projectile, is:
\[\Large {H_1} = - 16 \cdot {1.875^2} + 60 \cdot 1.875 + 100 = ...meters\]
please complete

- Michele_Laino

what is H_1?

- anonymous

idk

- Michele_Laino

try to compute it, using a calculator, for example

- anonymous

156.25

- Michele_Laino

correct!
\[\large H\left( t \right) = - 16 \cdot 1.8752 + 60 \cdot 1.875 + 100 = 156.25\;meters\]

- anonymous

Awesome is that it am I done????

- Michele_Laino

oops..
\[\large {H_1} = - 16 \cdot {1.875^2} + 60 \cdot 1.875 + 100 = 156.25\;meters\]

- Michele_Laino

now, we have to compute the time needed to our projectile, in order to reach the ground, starting from its position at H_1 with speed=0

- Michele_Laino

we have to use this equation:
\[\Large H\left( t \right) = - 16{t^2} + 156.25\]

- Michele_Laino

If I call with t_2 the time needed to our projectile in order to reach the earth surface, then we can write:
\[\Large 0 = - 16t_2^2 + 156.25\]
sinc qt t_2 H(t)=0
so the time t_2 is:
\[\Large {t_2} = \sqrt {\frac{{156.25}}{{15}}} = ...\sec \]
please compute that time t_2

- Michele_Laino

oops.. since at t_2 H(t)=0...

- Michele_Laino

oops.. another typo, here is the right formula:
\[\Large {t_2} = \sqrt {\frac{{156.25}}{{16}}} = ...\sec \]

- Michele_Laino

what is t_2?

- anonymous

3.125

- Michele_Laino

correct! So total time, needed to our projectile, is:
\[\Large T = {t_1} + {t_2} = 1.875 + 3.125 = ...\sec \]

- anonymous

5seconds

- Michele_Laino

ok!
more formally, if I call with h_0 the initial height, namely h_0=100 feet, and v_0 the initial speed, namely v_0=60 feet/sec, then time t_1 is:
\[\large {t_1} = \frac{{{v_0}}}{g}\]
as you can check from the above computations, furthermore, the maximum height reached by our projectile is:
\[\large {H_1} = \frac{{v_0^2}}{{2g}} + {h_0}\]
as you can check from the above computations again

- Michele_Laino

so time t_2 is:
\[\large {t_2} = \sqrt {\frac{{2{H_1}}}{g}} = \sqrt {\frac{2}{g}\left( {\frac{{v_0^2}}{{2g}} + {h_0}} \right)} \]

- Michele_Laino

and total time is:
\[\large T = {t_1} + {t_2} = \frac{{{v_0}}}{g} + \sqrt {\frac{2}{g}\left( {\frac{{v_0^2}}{{2g}} + {h_0}} \right)} \]
which represents the requested formula

- Michele_Laino

please, tell me when I may continue

- Michele_Laino

of course, g is gravity, namely g=32 feet/sec^2

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