Wednesday
March 29, 2017

Post a New Question

Posted by on .

Before working this problem, review Conceptual Example 15 . A pellet gun is fired straight downward from the edge of a cliff that is 22.3 m above the ground. The pellet strikes the ground with a speed of 34.9 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward?

  • Physics - ,

    h=v₀t +gt²/2
    v= v₀+gt
    v₀ =v-gt

    h=( v-gt)t + gt²/2 =
    =vt - gt² + gt²/2=
    = vt - gt²/2.
    gt² -2vt +2h = 0

    t = (2•34.9±sqrt{(2•34.9)²-4•9.8•2•34.9}/2•9.8 = 0.7 s
    v₀ =v-gt =34.9 – 9.8•0.7 = 28 m/s
    Upward motion
    H=v₀²/2g= 28²/2•9.8 = 40 m

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question