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.