Ask Question
11 September, 23:00

A ball is thrown downward from the top of a 190 -foot building with an initial velocity of 21 feet per second. the height of the ball h after t seconds is given by the equation h equals negative 16 t squared minus 21 t plus 190. how long after the ball is thrown will it strike the ground?

+3
Answers (1)
  1. 12 September, 00:01
    0
    To solve this problem you must apply the proccedure shown below:

    1 - You have the following information give in the problem above:

    - The ball is thrown downward from the top of a 190 -foot building with an initial velocity of 21 feet per second.

    - The height of the ball h after t seconds is given by this equation: h=-16t ²-21t+190.

    2 - h=0 when the ball strikes the ground, then you have:

    -16t²-21t+190=0

    3. When you solve the quadratic equation shown above, you obtain:

    t=2.85 seconds

    The answer is: 2.85 seconds.
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A ball is thrown downward from the top of a 190 -foot building with an initial velocity of 21 feet per second. the height of the ball h ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers