Animated Solution for Physics - Kinematics: From a place on the ground that is 20 m away from a wall, a bullet is fired aiming at a 50 m high mark on the wall. The location of the bullet is shown by a circular dot at some point of time. Where on the wall will the bullet hit?

Aiming High

Imagine you are a marksman standing $20\text{ m}$ away from a towering wall. Your target is a specific mark painted exactly $50\text{ m}$ high on that wall. When you align your sights, you establish a direct, straight line of sight to the target. This line represents the path the bullet would take if we lived in a universe without gravity.

From this geometry, we can immediately determine the initial angle of projection, $\theta$. Using simple trigonometry, the tangent of this angle is the ratio of the vertical height to the horizontal distance:

Gravity's Pull

However, we live in a world governed by gravity. The moment the bullet leaves the barrel, gravity begins to pull it downwards, causing it to deviate from that perfect straight line. The actual path of the bullet is a parabola, described by the standard equation of trajectory:

Notice the elegance of this equation. The first term, $x\tan \theta$, is exactly the height of our straight line of sight. The second term is the gravity drop—the exact distance the bullet falls below the line of sight due to gravity.

To find out where the bullet actually hits the wall, we need to know more about its initial velocity. Fortunately, the problem provides a graph showing a snapshot of the bullet's location. By carefully reading the grid, we can pinpoint this location.

The horizontal axis has major markings every $5$ units, with $5$ subdivisions, meaning each small square is $1\text{ m}$ wide. The vertical axis has major markings every $25$ units, with $5subdivisions$, meaning each small square is $5\text{ m}$ high. The dot is located at the second major horizontal tick ($x=10\text{ m}$) and exactly one small square below the $25\text{ m}$ mark ($y=20\text{ m}$).

So, we know that at $x=10\text{ m}$, the bullet is at a height of $y=20\text{ m}$.

Now, we substitute these coordinates into our trajectory equation to solve for the unknown velocity terms.

Rearranging this gives us the value of the entire gravity coefficient:

We don't need to isolate the initial velocity $u$; this combined constant is exactly what we need for the next step.

Finally, we want to find the height $y$ where the bullet strikes the wall, which is located at $x=20\text{ m}$. We plug $x=20$ and our newly found constant back into the trajectory equation:

The bullet hits the wall at exactly the $30\text{ m}$ mark.

This result beautifully illustrates the concept of gravity drop. If there were no gravity, the bullet would have hit the $50\text{ m}$ mark. During its flight to the wall, gravity pulled it down by exactly $20\text{ m}$!