An astronaut on the moon throws a baseball upward with an initial velocity of 10 meters per second, letting go of the baseball 2 meters abov

Question

An astronaut on the moon throws a baseball upward with an initial velocity of 10 meters per second, letting go of the baseball 2 meters above the ground. The equation of the baseball pathway can be modeled by h= -0.8t^2+10t+2. The same experiment is done on earth, in which the pathway is modeled by equation h= -4.9t^2+10t+2. How much longer would the ball stay in the air on the moon compared to on the Earth

in progress 0
Raelynn 2 months 2021-10-05T22:01:31+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-10-05T22:03:13+00:00

    Answer:

    The baseball will stay 10.47s longer on the moon than on the earth.

    Step-by-step explanation:

    The amount of time the baseball on the moon will stay  in air is until

    h(t)= -0.8t^2+10t+2 =0. (i.e when the ball reaches the ground)

    Similarly, the amount of time the baseball on earth will stay in air is until

    h(t)-4.9t^2+10t+2=0

    The solution to these equations can be found using the quadratic formula.

    For the baseball on the moon

    -0.8t^2+10t+2 =0

    t = \dfrac{-10\pm \sqrt{10^2-4(-0.8*2)} }{2*-0.8}

    whose positive solution is

    \boxed{t= 12.697s}

    And for the baseball on earth

    -4.9t^2+10t+2 =0

    t = \dfrac{-10\pm \sqrt{10^2-4(-4.9*2)} }{2*-4.9}

    whose positive solution is

    \boxed{t = 2.224s}

    Thus, the baseball will stay 12.697s-2.2243s=10.472s longer on the moon than on the earth.

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )