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physics

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a ball is thrown horizontally from a hill 29.0m high at a velocity of 4.00m/s. find the distance between the base of the hill and the pointwhere the ball hits the ground.

  • physics - ,

    intial Vy=0 (since it's thrown horizontal) change in y = -29.0m Ay=-9.81 m/s^2
    find the time used by equate :
    change in Y= intial Vy times time +1/2 Ay times time squared so -29.0=1/2(-9.81)t^2
    t^2 = .91s so t=2.43s
    since time used for y is same as x
    change in x = 1/2 (intial Vx+ final Vx) times time
    in progectile motion ini Vx = fin Vx
    so change in x = time(2.43s) times intial vx (4.00m/s))

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