anonymous
  • anonymous
help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@xapproachesinfinity
anonymous
  • anonymous
@Michele_Laino
anonymous
  • anonymous
whats your question?

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anonymous
  • 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
  • anonymous
i wish i could help with this one but im not sure sorry :(
anonymous
  • anonymous
thanks for trying
Michele_Laino
  • 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
  • Michele_Laino
furthermore the equation for speed v(t) is: \[\Large v\left( t \right) = 60 - 32t\]
Michele_Laino
  • 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
  • Michele_Laino
solving that equation for t_1 we have: \[\Large {t_1} = \frac{{60}}{{32}} = ...\sec \]
Michele_Laino
  • Michele_Laino
please complete that computation
Michele_Laino
  • Michele_Laino
please wait...
anonymous
  • anonymous
no problem you are a lifesaver
Michele_Laino
  • Michele_Laino
ok! I'm here
Michele_Laino
  • Michele_Laino
we get: \[\Large {t_1} = \frac{{60}}{{32}} = 1.875\sec \] am I right?
anonymous
  • anonymous
sure
Michele_Laino
  • 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
  • Michele_Laino
what is H_1?
anonymous
  • anonymous
idk
Michele_Laino
  • Michele_Laino
try to compute it, using a calculator, for example
anonymous
  • anonymous
156.25
Michele_Laino
  • Michele_Laino
correct! \[\large H\left( t \right) = - 16 \cdot 1.8752 + 60 \cdot 1.875 + 100 = 156.25\;meters\]
anonymous
  • anonymous
Awesome is that it am I done????
Michele_Laino
  • Michele_Laino
oops.. \[\large {H_1} = - 16 \cdot {1.875^2} + 60 \cdot 1.875 + 100 = 156.25\;meters\]
Michele_Laino
  • 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
  • Michele_Laino
we have to use this equation: \[\Large H\left( t \right) = - 16{t^2} + 156.25\]
Michele_Laino
  • 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
  • Michele_Laino
oops.. since at t_2 H(t)=0...
Michele_Laino
  • Michele_Laino
oops.. another typo, here is the right formula: \[\Large {t_2} = \sqrt {\frac{{156.25}}{{16}}} = ...\sec \]
Michele_Laino
  • Michele_Laino
what is t_2?
anonymous
  • anonymous
3.125
Michele_Laino
  • Michele_Laino
correct! So total time, needed to our projectile, is: \[\Large T = {t_1} + {t_2} = 1.875 + 3.125 = ...\sec \]
anonymous
  • anonymous
5seconds
Michele_Laino
  • 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
  • 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
  • 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
  • Michele_Laino
please, tell me when I may continue
Michele_Laino
  • Michele_Laino
of course, g is gravity, namely g=32 feet/sec^2

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