If a ball is thrown into the air with a velocity of 10m/s, its height in meters t seconds later is given by y=10t-4.9t^2.?

Update:

Find the average velocity for the time period beginning when t=1.5 and lasting 0.5 seconds.

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  • 6 months ago
    Favorite Answer

    At time t = 1.5 s, the height is 3.975 meters. At time t = 2.0 seconds, the height is 0.40 m. Average velocity = (change in height) / (change in time)= (-3.575 ) / (0.5) = -7.15 m/s.

  • 6 months ago

    The following equation is used to determine the height of the ball at a specific time.

    y = vi * t – ½ * a * t^2

    vi is 10 m/s. In this equation, -½ * a is equal to -4.9. So, the acceleration is -9.8 m/s^2. We can use the following equation to determine the ball’s vertical velocity at a specific time.

    vf = 10 – 9.8 * t

    At 1.5 seconds, vf = 10 – 9.8 * 1.5 = -4.7 m/s

    The negative sign means the ball is falling. According to the update, the acceleration lasts for 0.5 seconds. Let’s use the same equation to determine the ball’s velocity at 2 seconds.

    vf = 10 – 9.8 * 2 = -9.6 m/s

    Since the ball has a constant acceleration, we can use the following equation to determine its average velocity.

    Average velocity = ½ * (vi + vf)

    Average velocity = ½ * (-4.7 + -9.6) = -7.15 m/s

    I hope this is helpful for you.

  • oubaas
    Lv 7
    6 months ago

    Vav = ((10-g*1.5)+(10-g*2))/2 = (-5-10)/2 = -7.5 m/sec

  • 6 months ago

    I assume it is thrown straight upwards.

    y = 10t - 4.9t²

    y(1.5) = 15 – 4.9(1.5)² = 3.975 m

    y(2) = 20 – 4.9•4 = 0.4 m

    average velocity is usually listed as Δy/Δt or

    V = (3.975 – 0.4) / 1 = 3.575 m/s

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  • 6 months ago

    y = 10t - 4.9t²

    Average velocity is the change in position over change in time:

    v = (y(2.0) - y(1.5)) / (2.0 - 1.5)

    v = ((20 - 19.6) - (15 - 11.025)) / (0.5)

    v = (0.4 - 3.975) / 0.5

    v = -7.15

    The average velocity between t=1.5 and t=2.0 is -7.15 m/s.

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