Looking at the given answer choices, we can guess that the people who devised the question would choose (A) as its correct answer. Although (A) is better than the other three, it requires, to be correct,
at least one assumption, which is not mentioned in the question. Therefore, if we critically evaluate the question—and we should critically evaluate any mathematical problem and not make any unwarranted assumption—we find that there is no correct solution to the problem. Let us elaborate:
Consider Tarzan’s case first. When he lets go off the branch, he is falling under the influence of gravity ‘in a straight-line vertically downward’ with respect to the ground. Since his
initial velocity is 0, Tarzan’s flight is described by the following equation:
$h = 0.t + ½ gt^2 = 4.9t^2$ ... (1);
h is the distance measured from the Tarzan’s tree branch.
So, after 1 second, Tarzan falls 4.9 meter below the tree branch. In a piece paper, try to draw the points where Tarzan would be after 2, 3, … seconds.
Now what happens to Clayton’s bullet? When the bullet leaves the shotgun, it goes in a forward motion due to the bullet’s initial forward velocity. At the same time, it
simultaneously goes downward because of gravity. Both forward motion and downward motion are independent of each other and the distances that they cover in one unit of time, (say, 1 sec) are
different. Thus, the bullet's path will be a continuously curved line, i.e. a
parabola, not a diagonally straight line from Clayton's branch. You already know this if you have learnt the equations for
forward/horizontal distance and
downward/vertical distance, which are as follows:
$x = vt$ ... (2);
v is the initial horizontal velocity of the bullet.
v does not change over time as there is
no force in the forward direction.
$y = 0.t + ½ gt^2 = 4.9t^2$ ... (3); since the bullet’s
initial vertical velocity is 0, just like Tarzan.
(
x, y) are measured from Clayton's branch.
In a piece of paper, try to draw the points where the bullet will be after 2, 3, … seconds. Assume a value for
v, and first go along
x, then along
y, to plot the bullet's various positions.
Thus, after 1 sec, the bullet goes
x = vt = v.1 meter forward toward Tarzan and
y = 4.9 meter downward toward the ground.
So, you see that both Tarzan and the bullet are on the
same horizontal level after 1 second, i.e. both are
4.9 meter down from their respective branches. Now
if the tree branch is only 4.9 meter high, then the bullet will hit Tarzan only if
v.1 meter ≥ the horizontal distance (let, it be
d meter) between Tarzan and Clayton.
If
v.1 < d, the bullet and Tarzan both fall to the ground after exactly 1 sec, but the bullet misses Tarzan.
If
v.1 = d, the bullet hits Tarzan just at the ground after exactly 1 sec.
If
v.1 > d, the bullet hits Tarzan above the ground and before 1 sec (because at each instant, Tarzan and the bullet are always at the same horizontal level.
This is the key to understanding the question, which we would realize more below.)
Now
say that the branch is much higher than 4.9 meter and the bullet, although closer to Tarzan than before, could not hit him in the 1st second. But it has still chance because their flights have not finished. Let's see what happens after 2 seconds.
Tarzan falls 19.6 meter from his branch.
The bullet goes
v.2 meter forward toward Tarzan and 19.6 meter downward. Both the bullet and Tarzan are again on the same horizontal level. Now, if the branch is total 19.6 meter high, the bullet will hit Tarzan only if
v.2 ≥ d, as you can guess.
At this point, the whole picture should be clear to you.
(1) at each instant, $t = ..., ¼, ½, ¾, 1, 1.5, 2, ...$ sec, whatever it might be, the bullet and Tarzan are at the
same horizontal level.
(2) to be able to hit Tarzan in any of those levels, the
horizontal distance covered by the bullet at that level must be equal to the fixed distance between Tarzan and Clayton. This can be ensured if
v.t ≥ d.
(3) now the
maximum horizontal distance that the bullet can cover in its entire flight before reaching the ground is called its
range R. Thus, if
R ≥ d, the bullet will catch Tarzan sooner or later. This is the
implicit assumption made by the people who created the question. But
R depends on both bullet speed
v and branch height
H, and then R needs to be compared with
d. In real life, this assumption can be a good assumption given the high speed of shotgun bullets and a reasonable distance between the hunter and prey, if we forget air drag on the bullet.
But why should we assume R ≥ d in this theoretical problem without further information!
Therefore, the question has no precise answer.
Note: a picture is worth a thousand words. If you draw the graphs I suggested, everything would be clear to you quickly.