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anonymous
 one year ago
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anonymous
 one year ago
help

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@xapproachesinfinity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0whats your question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.0i wish i could help with this one but im not sure sorry :(

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1using your initial conditions, we can rewrite your equation, like below: \[\Large H\left( t \right) =  16{t^2} + 60t + 100\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1furthermore the equation for speed v(t) is: \[\Large v\left( t \right) = 60  32t\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1when 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
 one year ago
Best ResponseYou've already chosen the best response.1solving that equation for t_1 we have: \[\Large {t_1} = \frac{{60}}{{32}} = ...\sec \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please complete that computation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no problem you are a lifesaver

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we get: \[\Large {t_1} = \frac{{60}}{{32}} = 1.875\sec \] am I right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now 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
 one year ago
Best ResponseYou've already chosen the best response.1try to compute it, using a calculator, for example

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1correct! \[\large H\left( t \right) =  16 \cdot 1.8752 + 60 \cdot 1.875 + 100 = 156.25\;meters\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Awesome is that it am I done????

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1oops.. \[\large {H_1} =  16 \cdot {1.875^2} + 60 \cdot 1.875 + 100 = 156.25\;meters\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, 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
 one year ago
Best ResponseYou've already chosen the best response.1we have to use this equation: \[\Large H\left( t \right) =  16{t^2} + 156.25\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1If 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
 one year ago
Best ResponseYou've already chosen the best response.1oops.. since at t_2 H(t)=0...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1oops.. another typo, here is the right formula: \[\Large {t_2} = \sqrt {\frac{{156.25}}{{16}}} = ...\sec \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1correct! So total time, needed to our projectile, is: \[\Large T = {t_1} + {t_2} = 1.875 + 3.125 = ...\sec \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! 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
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.1and 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
 one year ago
Best ResponseYou've already chosen the best response.1please, tell me when I may continue

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1of course, g is gravity, namely g=32 feet/sec^2
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