If a ball is thrown straight up into the air with an initial velocity of 45 ft/s, it’s height in feet after t second is given by y

If a ball is thrown straight up into the air with an initial velocity of 45 ft/s, it’s height in feet after t second is given by y = 45t – 16t^2
a] find the average velocity for the time period begining when t = 2 and lasting.
i) 0.1 seconds
ii) 0.001 seconds
iii) 0.0001 seconds

b) Estimate the instantaneous velocity of the ball when t = 2

1 thought on “If a ball is thrown straight up into the air with an initial velocity of 45 ft/s, it’s height in feet after t second is given by y”

  1. The average velocity for the time period beginning when t = 2 and lasting is:  -19.016 ft/sec. The instantaneous velocity of the ball when t = 2 is: 19 ft/sec.

    Average velocity

    Velocity = 45-32t
    When t=2
    v(2) = 45-32(2)
    v(2) =45-64
    v(2) = -19 ft per second ( downward direction)
    Velocity=45-32t
    When t=2.1
    v(2.1) = 45-32(2.1)
    v(2.1) = 45-67.2
    v(2.1) = -22.2 ft per second
    Average velocity from t=2 to t=2.1
    Average velocity= (-19-22.2)/2
    Average velocity= -41.2/2
    Average velocity=-20.5 ft/sec
    Velocity=45-32t
    When t=2.001
    v(2.001) = 45-32(2.001)
    v(2.001) = 45-64.032
    v(2.001)  = -19.032.
    Average velocity from t=2 to 2.001
    Average velocity = (-19-19.032)/2
    Average velocity  = -19.016 ft/sec
    Velocity=45-32t
    When t=2.0001
    v(2.0001) = 45-32(2.0001)
    v(2.0001) =45-64.0032
    v(2.0001)  = -19.0032
    Average velocity from t=2 to 2.0001
    Average velocity= (-19-19.0032)/2
    Average velocity = -19.0016 ft/sec
    b. Instantaneous v at t=2
    Instantaneous = -19 ft/sec
    Therefore The average velocity for the time period beginning when t = 2 and lasting is:  -19.016 ft/sec. The instantaneous velocity of the ball when t = 2 is: 19 ft/sec.
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