March 28, 2017

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A projectile is launched straight up from the ground with an initial velocity of 320 ft/sec.

a. When will the projectiles height above the ground be 1024 ft?

For this I got 16 seconds

b. When will the height of the projectile be at least 24 ft?

I am not sure how to figure this one out.

c. When will the object strike the ground?

I got 20 seconds

d. What is the maximum height of the projectile?

I got 1600 feet.

So, I am still unsure about b and I wanted to know if my other answers seem correct


  • Pre-Calculus - ,

    c) and d) are correct

    for a) and the others I assume you used Calculus and the equation
    distance = -16t^2 + 320t to find your answers
    you must have set 1024 = -16t^2 + 320t
    after re-arranging and dividing by 16 this would give you
    t^2 - 20t + 64 = 0
    (t-16)(t-4) = 0
    so t = 16 or t = 4

    you had the t=16, but failed to show the t=4 case.
    It reached a height of 1024 twice, at 4 seconds after liftoff on its way up, and then 16 seconds later on its way down

    for b)
    set 24 = -16t^2 + 320t
    16t^2 - 320t + 24 = 0
    2t^2 - 40t + 3 = 0
    using the quadratic formula
    t = (40 ± √(1600-4(2)(3))/4
    = 19.92 or .075

    so between the times of .075 and 19.92 seconds the object is at least 24 feet high.

  • Pre-Calculus - ,

    My mistake. b should have asked at least 1024 ft. So does that mean the answer would be anytime in between 4 and 16 seconds?

    b. When will the height of the projectile be at least 1024 ft?

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