Question

Use the formula $h=-16t^2+v_{0}t$discussed in Example 9.

How fast would a ball have to be thrown upward to reach a maximum height of 100 ft? [Hint: Use the discriminant of the equation $16t^2-v_{0}t+h=0$]

Short answer

The ball would have to be thrown upwards with a speed of 80 ft/s to reach a maximum height of 100 ft.

Step-by-step solution

Step 1. State the concept.

An algebraic equation can be solved for the value of a variable by substituting the required values of all other variables in the equation.

Step 2. List the given data.

A ball follows a path described by the equation $h=-16t^2+v_{0}t$, where$h$ is the height of the ball above ground (in feet) at time$t$ seconds and$v_0$ is the initial speed of the ball.

The objective is to find the initial speed of the ball such that the maximum height reached by the ball is 100 ft.

The greatest height is the only height attained by the ball once. So, it is the value of$h$ for which the given equation would have real equal roots, that is, for which the discriminant is zero.

Step 3. Calculate the time.

Put $h=100$in $h=-16t^2+v_{0}t$to get,

$\begin{array}{l}100=-16t^2+v_{0}t\\ 16t^2-v_{0}t+100=0\end{array}$

Evaluate the discriminant of this equation to get,

role="math" localid="1643970871057" $$\begin{gathered} D={\left(-v_0\right)}^2-4\left(16\right)\left(100\right) \\ ={v_0}^2-6400 \end{gathered}$$

So, $D={v_0}^2-6400$.

Put $D=0$in $D={v_0}^2-6400$to get,

role="math" localid="1643970987764" $\begin{array}{l}0={v_0}^2-6400\\ {v_0}^2=6400\\ v_0=\sqrt{6400}\\ v_0=80\end{array}$

So, $v_0=80$. Note that only the positive square root is considered as speed is always non-negative.

This implies that a ball would have to be thrown upwards with a speed of 80 ft/s to reach a maximum height of 100 ft.