# Throwing a Ball

Alignments to Content Standards: A-CED.A.2

## Task

A ball thrown vertically upward at a speed of $v$ ft/sec rises a distance $d$ feet in $t$ seconds, given by $$d = 6 + vt - 16t^2$$

Write an equation whose solution is

1. The time it takes a ball thrown at a speed of 88 ft/sec to rise 20 feet.

2. The speed with which the ball must be thrown to rise 20 feet in 2 seconds.

## IM Commentary

Although this task is quite straightforward, it has a couple of aspects designed to encourage students to attend to the structure of the equation and the meaning of the variables in it:

• By focusing on different variables in parts (a) and (b), it fosters flexibility in seeing the same equation in two different ways: first as an equation in t with constants v and d, then as an equation in v with constants t and d.
• It does not give the values of the constants by means of an explicit equation such as $v=88$ or $d=20$. Thus it requires students to attend to the meaning of the variables in the preamble and extract the values from the descriptions in parts (a) and (b).

(Task from Algebra: Form and Function, McCallum et al., Wiley 2010)

## Solution

We have an equation in three unknowns, and each part specifies two of them giving an equation in one unknown whose solution gives the desired quantity.

1. We want $d = 20$, and we are given $v = 88$, so the equation is $20 = 6 + 88t - 16 t^2$.

2. We want $d = 20$, and we are given $t = 2$, so the equation is $20 = 6 + 2v - 16\cdot 2^2$. Simplifying the right-hand side, we get $20 = -58 + 2v$.