Question

question 10
a ball is launched from a 241.08-meter tall platform. the equation for the balls height ( h ) at time ( t ) seconds after launch is ( h = -4.9t^2 + 10.78t + 241.08 ), where ( h ) is in meters. what is the maximum height the ball achieves before landing? round your answer to two decimal places if necessary.

Explanation:

Step1: Find time of max height

For quadratic $h(t)=at^2+bt+c$, max time is $t=-\frac{b}{2a}$.
Here $a=-4.9$, $b=10.78$, so:
$$t=-\frac{10.78}{2\times(-4.9)} = \frac{10.78}{9.8} = 1.1$$

Step2: Compute max height

Substitute $t=1.1$ into $h(t)$.

$$\begin{align*} h(1.1)&=-4.9(1.1)^2 + 10.78(1.1) + 241.08\\ &=-4.9(1.21) + 11.858 + 241.08\\ &=-5.929 + 11.858 + 241.08\\ &=5.929 + 241.08\\ &=247.009 \end{align*}$$

Step3: Round to 2 decimals

Round $247.009$ to two decimal places: $247.01$

Answer:

247.01 meters