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Math Help - [SOLVED] got 5 problems i could use some help with

  1. #1
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    [SOLVED] got 5 problems i could use some help with

    1. An open box is to be constructed from a square sheet of plastic by removing a square of side 2 inches from each corner, and then turning up the sides. if the box must have a volume of 242 cubic inches, find the length of one side of the open box

    2.A ball is thrown vertically upward from the top of a building 112ft tall with an initial velocity of 96 feet per sec. the distance s (in feet) of the ball from the ground after t seconds is s=112+96t-16t^2.
    After how many seconds will the ball pass the top of the building on its way down?

    3. A projectile is fired from a cliff 400ft above the water at an inclination of 45degrees to the horizontal, with a muzzle velocity of 320 feet per second. the height h of the projectile above the water is given by
    h(x)=-32x^2/(320)^2+x+400, where x is the horizontal distance of the projectile from the base of the cliff. how far from the base of the cliff is the height of the projectile a maximum?

    4.Consider the quadratic model h(t)=-16t^2+40t+50 for the height (in feet), h of an object t seconds after the object has been projected straight up into the air. find the maximum height attained by the object. How much time does it take to fall back to the ground? Assume that it takes the same time for going up and coming down.

    5.A flare fired from the bottom of a gorge is visible when the lare is above the rim. If it is fired with an initial velocity of 176ft/sec, and the gorge is 480 ft deep, during what interval can the flare be seen? (h=-16t^2+vot+ho)
    the o's in the above formula are subscripts.

    Im sure 3 of the five all use the same formula because i can see their working is very similar as well as their functions. A little help please.
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  2. #2
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    Quote Originally Posted by ninobrn99 View Post
    1. An open box is to be constructed from a square sheet of plastic by removing a square of side 2 inches from each corner, and then turning up the sides. if the box must have a volume of 242 cubic inches, find the length of one side of the open box

    2.A ball is thrown vertically upward from the top of a building 112ft tall with an initial velocity of 96 feet per sec. the distance s (in feet) of the ball from the ground after t seconds is s=112+96t-16t^2.
    After how many seconds will the ball pass the top of the building on its way down?

    3. A projectile is fired from a cliff 400ft above the water at an inclination of 45degrees to the horizontal, with a muzzle velocity of 320 feet per second. the height h of the projectile above the water is given by
    h(x)=-32x^2/(320)^2+x+400, where x is the horizontal distance of the projectile from the base of the cliff. how far from the base of the cliff is the height of the projectile a maximum?

    4.Consider the quadratic model h(t)=-16t^2+40t+50 for the height (in feet), h of an object t seconds after the object has been projected straight up into the air. find the maximum height attained by the object. How much time does it take to fall back to the ground? Assume that it takes the same time for going up and coming down.

    5.A flare fired from the bottom of a gorge is visible when the lare is above the rim. If it is fired with an initial velocity of 176ft/sec, and the gorge is 480 ft deep, during what interval can the flare be seen? (h=-16t^2+vot+ho)
    the o's in the above formula are subscripts.

    Im sure 3 of the five all use the same formula because i can see their working is very similar as well as their functions. A little help please.
    1. Solve 242 = 2(x - 2)^2.

    2. Solve 112 = 112 + 96t - 16t^2. The larger solution is the answer.

    3. Solve h = 400 for x. The height is a maximum halfway between these two values of x.

    4. h(0) = 50. So the ground is at h = 50 ....? Get the turning point of the parabola. h-coordinate is the max height. Solve h = 50 for t, the larger solution is the time to return to ground.

    5. Where is h meant to be measured from?
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  3. #3
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    Hello, ninobrn99!

    5.A flare fired from the bottom of a gorge is visible when the flare is above the rim.
    If it is fired with an initial velocity of 176 ft/sec, and the gorge is 480 ft deep,
    during what interval can the flare be seen? . h\:=\:-16t^2+v_ot+h_o

    Let ground-level be h = 0

    We are given: . v_o = 176,\;h_o = -480

    The height function is: . h(t) \:=\:-16t^2 + 176t - 480
    . .
    This is a down-opening parabola.

    The flare can be seen when h(t) \,>\,0
    . .
    When is the parabola "positive"?

    We have: . -16t^2 + 176t - 480 \:=\:0
    . .
    The flare is at ground-level.

    Divide by -16: . t^2 - 11x + 30 \:=\:0\quad\Rightarrow\quad (t-5)(t-6) \:=\:0 \quad\Rightarrow\quad t \:=\:5,6


    Therefore, the flare can be seen from t = 5 to t = 6.

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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    1. Solve 242 = 2(x - 2)^2.

    2. Solve 112 = 112 + 96t - 16t^2. The larger solution is the answer.

    3. Solve h = 400 for x. The height is a maximum halfway between these two values of x.

    4. h(0) = 50. So the ground is at h = 50 ....? Get the turning point of the parabola. h-coordinate is the max height. Solve h = 50 for t, the larger solution is the time to return to ground.

    5. Where is h meant to be measured from?
    mr fantastic,
    I appreciate the way you help me work the problem, not solve it for me! It really helps me understand it!

    Quote Originally Posted by Soroban View Post
    Hello, ninobrn99!


    Let ground-level be h = 0

    We are given: . v_o = 176,\;h_o = -480

    The height function is: . h(t) \:=\:-16t^2 + 176t - 480
    . .
    This is a down-opening parabola.

    The flare can be seen when h(t) \,>\,0
    . .
    When is the parabola "positive"?

    We have: . -16t^2 + 176t - 480 \:=\:0
    . .
    The flare is at ground-level.

    Divide by -16: . t^2 - 11x + 30 \:=\:0\quad\Rightarrow\quad (t-5)(t-6) \:=\:0 \quad\Rightarrow\quad t \:=\:5,6


    Therefore, the flare can be seen from t = 5 to t = 6.

    Soroban,
    Thank you also for the thorough explination. the step by step break down helped dramtically!
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  5. #5
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    would you guys mind checking my work on this problem?

    A suspension bridge has twin towers that are 1300 feet apart. Each tower extends 180 ft above the road surface. The cables are parabolic in shape and are suspended from the tops of the towers. The cables touch the road surface at the center of the bridge. Find the height of the cable at a point 200 feet from the center of the bridge.

    I came up with:
    \frac{180}{(650)^2}x^2

    \frac{180}{422500}(200)^2

    \frac{180}{422500}40000

    \frac{2880}{168}

    \sim17.041ft
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