Question #43d72

2 Answers
Aug 31, 2017

62.8 meters


If #x# is the height at any time t, then #x'' = (d^2x)/(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(t) = dx/dt = -9.81t + 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(t) = 12 - 9.8t = dx/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 - 9.8t = 0#
#12 = 9.8t#
#t = 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(t) = -12 - 9.8t#

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

#x(t) = -12t - (9.8/2)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(2.56) = 0 = (-12 * 2.56) - (4.9 * 2.56^2) + 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



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


This becomes:



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