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anonymous

  • one year ago

HELP!! A projectile is launched straight up at 62.5 m/s from a height of 73.5 m, at the edge of a sheer cliff. The projectile falls, just missing the cliff and hitting the ground below. (a) Find the maximum height of the projectile above the point of firing. (m) (b) Find the time it takes to hit the ground at the base of the cliff. (s) (c) Find its velocity at impact. (m/s)

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  1. anonymous
    • one year ago
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    a)maximum height , Hmax = v^2/(2 * g)

  2. anonymous
    • one year ago
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    a) maximum height = 62.5^2/(2 * 9.8) maximum height =199.3 m

  3. anonymous
    • one year ago
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    b) let the time taken to fall is t Using second equation of motion y = ut + 0.5 at^2 -73.5 = 62.5 * t - 0.5 * 9.8 t^2 solving for t t = 13.8 s the time taken to fall is 13.8 s

  4. anonymous
    • one year ago
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    part c) let the velocity of impact is v m/s using third equation of motion v^2 - u^2 = 2 * a * d

  5. anonymous
    • one year ago
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    v^2 - 62.5^2 = 2 * 9.8 * 73.5 solving for v v = 73.12 m/s the velocity of impact is 73.12 m/s

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