# Question #43d72

Aug 31, 2017

62.8 meters

#### Explanation:

If $x$ is the height at any time t, then $x ' ' = \frac{{d}^{2} x}{{\mathrm{dt}}^{2}}$

And at earth's surface this is a constant 9.81 meters per second per second. (rounding to 3 significant digits because that's what we did on our slide rules oh-so-many years ago.)

then $v \left(t\right) = \frac{\mathrm{dx}}{\mathrm{dt}} = - 9.81 t + c$

We're saying -9.81t because we're assigning negative values to the downward direction, positive values to the upward direction.

We can calculate the value of constant c because we know the value of v at time 0: 12m/s.

so we now know that $v \left(t\right) = 12 - 9.8 t = \frac{\mathrm{dx}}{\mathrm{dt}}$

We know the ball will travel upwards while decelerating, eventually reaching velocity 0 at the apex of its travel. We can solve for the time when this occurs:

$12 - 9.8 t = 0$
$12 = 9.8 t$
$t = \frac{12}{9.8} = 1.22$ seconds (rounding to 3 digits again)

We know that at time t = 2 * 1.22, or 2.44 seconds, the ball is back to it's original height, and is travelling downwards at 12 m/s. (We're ignoring wind resistance).

Since we're told that the ball's total flight time is 5 seconds, we know that it has $5 - 2.44 = 2.56$ seconds flight time remaining.

At this point, we can start a completely new derivation. We have known initial conditions v(0) = -12 m/s, and total elapsed time will be 2.56 seconds.

So we have velocity $v \left(t\right) = - 12 - 9.8 t$

...and we can integrate to find x(t):

$x \left(t\right) = - 12 t - \left(\frac{9.8}{2}\right) {t}^{2} + c$

...and we know that at time t = 2.56 the ball's height will be zero.

With this, we can solve for constant c, which will be the tower's height (and the ball's height at time t = 0).

$x \left(2.56\right) = 0 = \left(- 12 \cdot 2.56\right) - \left(4.9 \cdot {2.56}^{2}\right) + c$

$- 30.7 - 32.1 + c = 0$

c = 62.8 meters. (rounded to 3 significant digits in all calculations)

...and you can see that we sidestepped the need to calculate the max height that the ball reached.

Sep 1, 2017

$\textsf{63 \textcolor{w h i t e}{x} m}$

#### Explanation:

Setting the ball at the origin and using the convention "up is positive" we can use:

$\textsf{s = u t + \frac{1}{2} a {t}^{2}}$

This becomes:

$\textsf{s = \left(12 \times 5\right) - \left(\frac{1}{2} \times 9.81 \times {5}^{2}\right)}$

$\textsf{s = 60 - 122.625 = - 62.625 \textcolor{w h i t e}{x} m}$

This means the height of the tower can be rounded to 63 m.