Honors Algebra 2: Projectile Motion Extra Practice

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Recall that the height, in feet, of a projectile thrown upward from an initial height of h0 feet with an initial velocity of v0 feet per second is given by \( h(t) = -16t^2 + v_0 t + h_0\)

Simplify:

  1. A projectile is thrown up from the ground with an initial velocity of 48 feet per second.\( \displaystyle \)
    1. How high is the ball 1/3 of a second later? Answer to the nearest hundredth of a foot.
      2. Solution 14.22 feet
    2. When does the ball hit the ground?
      3. Solution 3 seconds
    3. When is the ball the highest?
      4. Solution 1.5 seconds
    4. What's the highest the ball reaches?
      5. Solution 36 feet
    5. When is the ball 50 feet high on the way down? Answer to the nearest tenth of a second
      6. Solution 2.8 seconds
  2. A projectile is thrown up from an initial height of 30 feet with an initial velocity of 120 feet per second.
    1. How high is the ball 3/4 of a second later?
      2. Solution 111 feet
    2. When does the ball hit the ground? Answer to the nearest tenth of a second.
      3. Solution 7.7 seconds
    3. When is the ball the highest? Answer to the nearest hundredth of a second
      4. Solution 3.75 seconds
    4. What's the highest the ball reaches?
      5. Solution 255 feet
    5. When is the ball 50 foot high on the way up? Answer to the nearest hundredth of a second
      6. Solution 0.17 seconds