Question

help

Answer 1

@xapproachesinfinity

Answer 2

@Michele_Laino

Answer 3

whats your question?

Answer 4

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)

Answer 5

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

Answer 6

thanks for trying

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

please complete that computation

Answer 12

please wait...

Answer 13

no problem you are a lifesaver

Answer 14

ok! I'm here

Answer 15

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

Answer 16

sure

Answer 17

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

Answer 18

what is H_1?

Answer 19

idk

Answer 20

try to compute it, using a calculator, for example

Answer 21

156.25

Answer 22

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

Answer 23

Awesome is that it am I done????

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

what is t_2?

Answer 31

3.125

Answer 32

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

Answer 33

5seconds

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

please, tell me when I may continue

Answer 38

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