Mathematico | Mathematico

A ball is dropped from the top of a building of height $63 m .$ At time $t$ seconds after being dropped, the ball has traveled a distance of

4.9 t^{2} m

At the same time the ball is dropped, an elevator at the bottom of the building starts to ascend at a constant speed of $0.7 {ms}^{- 1}$ .

If the ball and elevator pass each other at time $T$ seconds,

Show that
$7 T^{2} + T - 90 = 0$
Hence, use completing the square to show that
$T = \frac{\sqrt{k} - 1}{14}$
for some integer $k .$

Give the value $k$ .

Hint

The distance traveled by the elevator after $t$ seconds will be

\begin{aligned} distance & = speed \times time \\ = 0.7 t \end{aligned}

Hint

At the correct moment, the distances traveled by the ball and the elevator must add together to give the height of the building:

Ball falling from building

Solution

The distance traveled by the elevator after $t$ seconds will be

\begin{aligned} distance & = speed \times time \\ = 0.7 t \end{aligned}

At the correct moment, the distances traveled by the ball and the elevator must add together to give the height of the building:

Ball falling from building

So we get
$\begin{aligned} 4.9 T^{2} + 0.7 T & = 63 \\ 49 T^{2} + 7 T & = 630 \\ 7 T^{2} + T - 90 & = 0 \end{aligned}$
Solving for $T$
$\begin{aligned} T^{2} + \frac{1}{7} T - \frac{90}{7} & = 0 \\ {(T + \frac{1}{14})}^{2} - \frac{1}{196} - \frac{90}{7} & = 0 \\ {(T + \frac{1}{14})}^{2} & = \frac{2521}{196} \\ T & = - \frac{1}{14} \pm \sqrt{\frac{2521}{196}} \\ T & = \frac{\sqrt{2521} - 1}{4} \end{aligned}$
(Note we discarded a solution since clearly $T > 0$ .)