To find out when the ball will hit the ground, we first need to rewrite the height function \(h(t)\) and then find the time at which the ball reaches a height of 0 meters.
- The original function is given by: \[ h(t) = -5t^2 + 40t \]
To rewrite this function in a factored form, we can factor out the common terms:
\[ h(t) = -5(t^2 - 8t) \]
Next, we can complete the square inside the parentheses. The expression \(t^2 - 8t\) can be rewritten by completing the square:
\[ t^2 - 8t = (t - 4)^2 - 16 \]
So, substituting this back into the equation gives us: \[ h(t) = -5((t - 4)^2 - 16) \] \[ h(t) = -5(t - 4)^2 + 80 \]
Now we have the function in vertex form: \[ h(t) = -5(t - 4)^2 + 80 \]
The height \(h(t)\) reaches ground level (0 meters) when: \[ -5(t - 4)^2 + 80 = 0 \]
- To find when the ball hits the ground, we can set \(h(t) = 0\) and solve for \(t\):
Starting from the factored vertex form: \[ -5(t - 4)^2 + 80 = 0 \] Adding \(5(t - 4)^2\) to both sides yields: \[ 5(t - 4)^2 = 80 \] Dividing both sides by 5 gives: \[ (t - 4)^2 = 16 \] Taking the square root of both sides results in: \[ t - 4 = 4 \quad \text{or} \quad t - 4 = -4 \] Thus, solving for \(t\):
- \(t - 4 = 4 \Rightarrow t = 8\)
- \(t - 4 = -4 \Rightarrow t = 0\)
The ball is launched at \(t = 0\) and hits the ground at \(t = 8\) seconds.
Thus, the answers are:
-
Rewritten function: \[ h(t) = -5(t - 4)^2 + 80 \]
-
The ball hits the ground: \[ t = 8 \text{ seconds} \]