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5 questions
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  • Q1
    What are the variables given and implied from the situation above?
    Question Image
    d=-443m, a=0m/s^2, vi=9.8m/s
    d=9000m, a=-9.8m/s^2, vi=0m/s
    d=-443m, a=-9.8m/s^2, vi=0m/s
    d=-443m, a=-9.8m/s^2, vi=-10m/s
    120s
    112.39.c.4.b
  • Q2
    Considering these known variables, which of the four equations above is needed to determine how long the penny will take to reach the ground?
    Question Image
    a=(vf^2-vi^2)/2d
    d=vi*t+(1/2)a*t^2
    a=(vf-vi)/t
    v=d/t
    120s
    112.39.c.4.b
  • Q3
    How long will it take for the penny to strike the ground, assuming no air resistance?
    Question Image
    t=50.6s
    t=9.5s
    t=90.4s
    t=20.2s
    120s
    112.39.c.4.b
  • Q4
    How long would the penny take to strike the ground if the Empire State Building was twice as tall?
    Question Image
    t=20.5s
    t=13.4s
    t=93s
    t=35.3s
    120s
    112.39.c.4.b
  • Q5
    What would the acceleration vs. time graph look like for the penny?
    Question Image
    Flat line at 9.8m/s^2
    Straight line with a positive constant slope of 9.8m/s^2
    Straight line with a negative constant slope of -9.8m/s^2
    Flat line at -9.8m/s^2
    120s
    112.39.c.4.b

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