Heavenaubryidn
Answered

Explore a diverse range of topics and get answers from knowledgeable individuals on IDNLearn.com. Join our knowledgeable community and get detailed, reliable answers to all your questions.

Deon throws a ball into the air. The height, h, of the ball, in meters, at the time t seconds is modeled by the function h(t) = -5t + t + 4.

1. When will the ball hit the ground?

2. Will the ball reach a height of 5 meters?

Sagot :

Mhanifaidn

Answer:

  • 1. After 1 second,
  • 2. No.

Step-by-step explanation:

Given function

  • h(t) = - 5t² + t + 4

Solution

1) Ball hits the ground when:

  • h(t) = 0, so

  • - 5t² + t + 4 = 0
  • 5t² - t - 4 = 0
  • 5t² + 4t - 5t - 4 = 0
  • t(5t + 4) - (5t + 4) = 0
  • (t - 1)(5t + 4) = 0
  • t - 1 = 0 or 5t + 4 = 0
  • t = 1 or t = - 4/5

Solution is t = 1 and the second solution is discarded as negative.

The ball will hit the ground after 1 second.

2) Let's assume h(t) = 5, solve for t:

  • - 5t² + t + 4 = 5
  • - 5t² + t - 1 = 0
  • 5t² - t + 1 = 0

  • D = (-1)² - 4*5 = 1 - 20 = - 19

No solutions as the discriminant is negative, so the ball will not reach 5 meters.

Semsee45idn

Answer:

1.  1 second.

2.  No.

Step-by-step explanation:

Given function:

[tex]h(t)=-5t^2+t+4[/tex]

where:

  • h is the height of the ball (in meters).
  • t is the time (in seconds).

Question 1

The ball will hit the ground when the height is zero.

Substitute h = 0 into the given function and solve for t:

[tex]\begin{aligned}h(t)&=0\\\implies -5t^2+t+4&=0\\-1(5t^2-t-4)&=0\\5t^2-t-4&=0\\5t^2-5t+4t-4&=0\\5t(t-1)+4(t-1)&=0\\(5t+4)(t-1)&=0\\\\5t+4&=0\implies t=-0.8\\t-1&=0 \implies t=1\end{aligned}[/tex]

As t ≥ 0, t = 1 only.

Therefore, the ball will hit the ground after 1 second after it is thrown in the air.

Question 2

The x-coordinate of the vertex of a quadratic function is:

[tex]\boxed{x=-\dfrac{b}{2a}}\quad \textsf{when $y=ax^2+bx+c$}.[/tex]

Therefore, the x-coordinate of the vertex of the given function is:

[tex]\implies t=-\dfrac{1}{2(-5)}=\dfrac{1}{10}=0.1[/tex]

Substitute t = 0.1 into the function to find the maximum height of the ball:

[tex]\implies h(0.1)=-5(0.1)^2+0.1+4=4.05[/tex]

Therefore, the ball will not reach a height of 5 m, as the maximum height it can reach is 4.05 m, and 5 > 4.05.

Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com provides the best answers to your questions. Thank you for visiting, and come back soon for more helpful information.