One football is kicked into the air with an initial vertical velocity of 44 feet per second. Another football is kicked it to the air with a

Question

One football is kicked into the air with an initial vertical velocity of 44 feet per second. Another football is kicked it to the air with an initial vertical velocity of 40 feet per second.
a. Which football is in the air for more time?
b. Justify your answer to part (a).

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Trúc Chi 3 weeks 2021-08-23T23:27:52+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-08-23T23:28:58+00:00

    Answer:

    a. The football with initial vertical velocity 44 ft per second b. Time in air for 44 ft per second ball = 2.76 s . Time in air for 40 ft per second ball = 2.50 s

    Explanation:

    a. The football with initial vertical velocity 44 ft per second

    b. Using v = u + at where v = velocity at maximum height = 0

    For first football u = 44 ft/s,a = g = -32 ft/s²

    v = u + at

    0 = 44 + (-32)t

    0 = 44 -32t

    -44 = -32t

    t ₁= 44/32 = 1.38 s

    The vertical distance it moves is gotten from v² = u² + 2as with v = 0,

    s = u²/2a = 44²/(2 × 32) = 30.25 ft

    Since it covers this same distance on its downward fall, its velocity as it as it hits the ground is v² = u² + 2as where u = 0 and g = -32ft/s²

    v² = u² + 2as = 0 + 2 ×(-32) ×(-30.25) = 1936 ⇒ v =√1936 = 44 ft/s

    The time it takes to cover this distance is gotten from v = u + at with u = 0

    -44 = 0 + (-32)t

    -44 = 0 -32t

    -44 = -32t

    t₂ = 44/32 = 1.38 s

    total time = t₁ + t₂ = 1.38 s + 1.38 s = 2.76 s

    For second football u = 40 ft/s,a = g = -32 ft/s²

    v = u + at

    0 = 40 + (-32)t

    0 = 40 -32t

    -40 = -32t

    t₃ = 40/32 = 1.25 s

    The vertical distance it moves is gotten from v² = u² + 2as with v = 0,

    s = u²/2a = 40²/(2 × 32) = 25 ft

    Since it covers this same distance on its downward fall, its velocity as it as it hits the ground is v² = u² + 2as where u = 0 and g = -32ft/s²

    v² = u² + 2as = 0 + 2 ×(-32) ×(-25) = 1600 ⇒ v =√1600 = 40 ft/s

    The time it takes to cover this distance is gotten from v = u + at with u = 0

    -40 = 0 + (-32)t

    -40 = 0 -32t

    -40 = -32t

    t₄ = 40/32 = 1.25 s

    total time = t₃ + t₄ = 1.25 s + 1.25 s = 2.50 s

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