a tennis ball is thrown straight up at a speed of 40m/s and caught at the same level. calculate rhe maximum height reached by the ball​

Question

a tennis ball is thrown straight up at a speed of 40m/s and caught at the same level. calculate rhe maximum height reached by the ball​

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Ngọc Hoa 1 year 2021-09-03T15:54:40+00:00 2 Answers 28 views 0

Answers ( )

    0
    2021-09-03T15:55:50+00:00

    Answer:

    81.6 m

    Explanation:

    Answer: 81.6 m.

    The time it takes gravity to slow 40 m/s to zero when it teaches maximum height is

    -v(initial) / -g = t

    -40 m/s / -9.8 m/s^2 = 4.08 s

    The height reached is the average velocity times this time 4.08 s, with v(avg) = [v(initial) + v(final)] / 2 with v(final) = 0. v(avg) = v(initial) / 2 = 40 m/s / 2 = 20 m/s.

    So the distance d of maximum height is

    d = v(avg)•t

    d = 20 m/s • 4.08 s = 81.6 m.

    0
    2021-09-03T15:55:55+00:00

    Answer:

    80m

    Explanation:

    time taken for stone to reach max-height is t=final velocity – intial velocity / acceleration due to gravity

    t=v-u/g

    where:

    v=0 , u=40m/s ,g=-10m/so

    : t=0-40/-10

    t=4secs

    : max-height reach=ut + 0.5gt²

    =(40)(4)+(0.5)(-10)(4)I

    =160-5(16)

    =160-80

    =80m

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