Mathematics

Question

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?

2 Answer

  • 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.

  • 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.

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