Monkey on the tower problem

A monkey, standing on a tower, throws a banana.  The initial velocity of the banana is always exactly horizontal.

There are three variables in this problem that are either given, or to be solved for:  the height of the tower (d_y), the speed of the banana (v_x) and the distance over the ground that the banana travels before it hits the ground (d_x).

Air resistance is ignored, so that the only force on the banana after it leaves the monkey's hand is the downward force of gravity.  Since this is perpendicular to the horizontal velocity given to the banana by the monkey, gravity can not affect the horizontal speed of the banana.

Therefore, the horizontal velocity does not change throughout the problem, and it can be calculated from the definition of velocity.  v_x = d_x / t

The time that it takes for the banana to hit the ground does not depend on the speed that the monkey throws the banana.
This is true because the monkey throws the banana horizontally.  Gravity does not affect the horizontal speed and the horizontal speed does not affect the rate of fall.  These speeds are totally independent of each other because they are perpendicular to each other.  We have demonstrated this independence in class.

The height of the tower is related to the time that it takes the banana to fall.  d_y = ¹/₂ at²
Time (t) appears in both equations.  If the given information includes the height of the tower, we can always use d_y = ¹/₂ at²  to calculate the time required for the banana to fall.  If d_y  is the unknown, then we can use v_x = d_x / t   to calculate the time required for the banana to fall.  Either way, the first step to the "monkey on the tower" problem is to find the time to fall.

The second step is always to use the other (so far unused) equation to solve for the unknown.

Example:
If the monkey throws the banana off of a  tower that is 100 m tall, and the banana travels a horizontal distance of 90 m, how fast was the banana going when it left the monkey's hand?

Solution:
Since we know the height of the tower is 100 m, when can use this to calculate the time to fall.  Starting with d_y = ¹/₂ at²,  we can rearrange the formula to solve for time.  This yields t = \/(2d_y /a) = \/[2(100 m)/(9.81m/s²)] = 4.5 s

We now have everything we need to solve the second equation:  v_x = d_x / t   =  (90 m)/(4.5 s) = 20 m/s.
The monkey threw the banana horizontally with a speed of 20 m/s.
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Cannon Problem

A cannon fires a projectile with a given muzzle velocity (v_r), and a given angle of elevation (f).
We wish to find how far down range the projectile will fly (d_x).

As in the previous problem, gravity does not affect the horizontal speed, and v_x = d_x / t.
Rearranged to solve for d_x, we have the final step to this problem:  d_x =  v_x t
We could therefore calculate the distance if we knew the horizontal velocity and the time aloft.

The horizontal velocity stays the same throughout the flight of the projectile, and is the horizontal component of the muzzle velocity.  Thus we can calculate the horizontal velocity,  v_x = v_rcosf.

Now all we need to do is to calculate the time aloft (t_T).

We have learned in class and from our text that the right half of the projectile trajectory is symmetric with the left.  The total time aloft (t_T) is therefore twice the time to either rise or fall (t_1/2).  The time to fall, and the second half of the trajectory, is basically the same as the "monkey on the tower problem."

We have also learned in class that if the launcher and target are at the same level (altitude or height), the impact speed is the same as the launching (muzzle) speed.  The size of the y-components of the velocity are also the same at launch and impact (but opposite in direction).

This implies that the y part of the muzzle velocity, v_y = v_rsinf, is the same as the v_f  for an object dropped from the highest point of the trajectory!  This speed can be found from the ch. 2  formula,   v_f  = at.  The t is our time to fall, t_1/2 .
And:    v_f  = v_y .
So   v_y = at_1/2.   ==>  t_1/2 = v_y/a

We now are ready to make a 5-step "recipe" for finding the range of a projectile when the muzzle velocity and gun angle of elevation are known.

            1     v_x =  v_rcosf
            2        v_y = v_rsinf
            3        t_1/2= v_y/a
            4        t_T = 2t_1/2
            5         d_x =v_x t_T

Example:
If a gun has a muzzle velocity of 100 m/s and an angle of elevation of 30^o, find the range of the gun.

               1      v_x =  v_rcosf = (100 m/s)cos30^o  = (100 m/s)(0.8660) = 86.6 m/s.
               2        v_y = v_rsinf = (100 m/s)sin30^o = (100 m/s)(0.5000) = 50.0 m/s.
                3        t_1/2= v_y/a =  (50.0 m/s)/(9.81m/s²) = 5.10 s
                4        t_T = 2t_1/2  =   2(5.10 s) = 10.2 s
                5         d_x =v_x t_T =  (86.6 m/s)(10.2 s) = 880 m

The projectile traveled 880 m down range.
 

        Can these five steps be combined into a single formula?
        What is lost when you do?  Why didn't I teach it that way?