Taejm76idn
Answered

IDNLearn.com offers expert insights and community wisdom to answer your queries. Ask your questions and receive prompt, detailed answers from our experienced and knowledgeable community members.

Your friend tosses a ball into the air at an initial velocity of 18 feet per second. The equation h = -8t^2 + 18t + 5 models the height h of the ball t seconds after it was thrown.

1. How long was the ball in the air?

2. How high did the ball get?

3. When did the ball hit its highest point?​

Sagot :

Sqdancefanidn

Answer:

  1. 2.5 s
  2. 15.125 ft
  3. 1.125 s

Step-by-step explanation:

The time and height of the highest point represent the vertex of the graph of the height equation. The positive zero of the equation represents the time the ball was no longer in the air.

__

These points are conveniently found from the vertex form of the equation.

  h = -8(t² -9/4t) +5

  h = -8(t² -9/4t +81/64) +5 +8(81/64)

  h = -8(t -9/8)² +15 1/8 . . . . . . . . vertex is (9/8, 15 1/8)

__

Solving for the zeros, we find ...

  0 = -8(t -9/8)² +15 1/8

  121/64 = (t -9/8)² . . . . . . subtract 15 1/8, divide by -8

  11/8 = t -9/8 . . . . . . . take the positive square root

  t = 20/8 = 2.5 . . . . the time the ball hits the ground

__

1. The ball was in the air for 2.5 seconds

2. The ball got to a height of 15 1/8 feet

3. The ball was at its highest point 1.125 seconds after it was tossed.

View image Sqdancefan
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Discover the answers you need at IDNLearn.com. Thanks for visiting, and come back soon for more valuable insights.