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a long jump athlete take off at 25 degree with the horizontal and achieves a jumping distance of 9.12 m. Calculate initial take off speed.

  • physics -

    but time in air can be found by
    or 0=0+vy*t-4.9t^2
    0=t(vy-4.9) or time in air=vy/4.9

    but vy=vsin25 and vh=vcos25

    time in air=V*sin25/4.9
    9.12=Vcos25*Vsin25/4.9-4.9(V *sin 25/4.9)^2

    and you solve with V with some algebra and trig.

  • physics -


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