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 ba
Question
1. When will the ball hit the ground?
2. Will the ball reach a height of 5 meters?
2 Answer

1. User Answers mhanifa
Answer:
 1. After 1 second,
 2. No.
Stepbystep 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.

2. User Answers semsee45
Answer:
1. 1 second.
2. No.
Stepbystep 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^2t4)&=0\\5t^2t4&=0\\5t^25t+4t4&=0\\5t(t1)+4(t1)&=0\\(5t+4)(t1)&=0\\\\5t+4&=0\implies t=0.8\\t1&=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 xcoordinate 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 xcoordinate 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.