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16 February, 13:11

An arrow is shot vertically upward from a platform 16ft high at a rate of 190ft/sec. When will the arrow hit the ground? Use the formula: h=-16t²+vt+h0. (Round your answer to the nearest tenth.)

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  1. 16 February, 13:12
    0
    H (t) = - 16t2 + 186t + 43

    at the ground h = 0

    hence; - 16t2 + 186t + 43 = 0

    solving this quadratic equation using the quadratic formula; a = - 16, b = 186, c = 43; x = (-b + - (b2 - 4ac) 1/2) / 2a

    gives t = 11.8 seconds to the nearest tenth (note that the negative root has no practical significance)
  2. 16 February, 13:35
    0
    time required for arrow to reach ground is 10.8 sec

    when t = 10.8 seconds to the nearest tenth

    Step-by-step explanation:

    Given values are

    Velocity = 190 ft/sec height = 16 ft

    Given formula is

    h=-16t²+vt+h0

    adding the values, we get

    h (t) = - 16t²+190t+16

    so we have to find when the hit he ground, so at ground the height will be 0

    0 = - 16t²+190t+16

    -16t²² + 190t + 16 = 0

    using the quadratic formula

    x = (-b + (b2 - 4ac) 1/2) / 2a

    values

    a = - 16, b = 190, c = 16

    x = - 190 + ((-190) ² - 4 (-16) (16) ½ / 2 (-16)

    x = - 190 + ((-190) ²+1024) ½ / 2 (-16)

    x = - 190 + (-190² + 32²) ½ / -32

    taking under root

    x = - 190 + (-190+32) / - 32

    x = 190 + (-158) / -32

    we have 2 options,

    x = 190 + (-158) / -32 x = 190 - (-158) / -32

    x = 190 - 158) / -32 x = 190 + 158) / -32

    x = - 1 x = 10.875

    or t = 10.875

    so the negative root has no practical significance

    and we have t = 10.8 seconds to the nearest tenth
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