Animated Solution for Physics - Kinematics: A particle is thrown with a speed $v$ from a point $O$ at an angle $\theta$ with the horizontal plane such that it passes through the point $P$ at a height of $1\text{ m}$ and horizontal distance of $5\text{ m}$ from $O$, as shown in the figure. If acceleration due to gravity is $g{\text{ ms}}^{-2}$, then the correct statement (s) is/are :

Imagine you are standing at the origin of a coordinate system, holding a particle. Your goal is to throw this particle so that it perfectly passes through a specific target in the air: point $P$.

We are given that point $P$ is located $5\text{ m}$ horizontally and $1\text{ m}$ vertically from our starting position. This means the coordinates of our target are $(5,1)$.

To solve this, we need a mathematical tool that describes the path of the particle through space.

The most powerful tool for this job is the equation of trajectory. Unlike the standard equations of motion that depend on time, the trajectory equation directly relates the vertical height $y$ to the horizontal distance $x$.

Since we know the particle must pass through $P(5,1)$, we can substitute $x=5$ and $y=1$ into our master equation.

Simplifying this gives us a raw, fundamental relationship between the launch angle $\theta$ and the initial speed $v$:

Now, let's evaluate the given options systematically. Option A suggests a launch angle of $\theta =45^{\circ }$. Let's plug this into our relationship.

We know that $\tan 45^{\circ }=1$ and $\cos 45^{\circ }=\frac{1}{\sqrt{2}}$. Substituting these values:

Rearranging the terms to solve for $v^2$:

Taking the square root gives us the required velocity:

This perfectly matches the value given in Option A! Therefore, Option A is correct.

But what about the particle's maximum height? Option B claims it reaches its peak before hitting point $P$. To verify this, we need to calculate the total horizontal range $R$.

Substituting our velocity and $\theta =45^{\circ }$:

Because projectile motion is perfectly symmetric, the maximum height occurs exactly at the halfway mark of the range.

Since $3.125\text{ m}$ is strictly less than our target distance of $5\text{ m}$, the particle reaches its peak and begins to descend before it ever reaches point $P$. Thus, Option B is also correct.

Let's shift our focus to Option C, which proposes a launch angle of $\theta =30^{\circ }$. To analyze this, we can use an elegant alternate form of the trajectory equation that incorporates the range $R$:

Substituting our coordinates $(5,1)$ and $\theta =30^{\circ }$:

Solving for the range $R$:

Once again, the maximum height occurs at half the range:

Just like the 45-degree scenario, $3.825\text{ m}$ is less than $5\text{ m}$. The particle reaches its maximum height before reaching $P$. Therefore, the claim in Option C is incorrect.

Finally, let's examine Option D. It suggests a launch angle where $\theta ={\tan }^{-1}\left(\frac{1}{5}\right)$. This implies that $\tan \theta =\frac{1}{5}$.

Let's substitute this back into our fundamental relationship:

Subtracting 1 from both sides leaves us with a fascinating mathematical situation:

For a fraction with a non-zero numerator to equal zero, the denominator must approach infinity. This means the velocity $v$ must be infinite ($v\to \infty$).

Physically, this makes perfect sense. If you aim directly at the target (since the angle to the target is exactly ${\tan }^{-1}(1/5)$), gravity will immediately start pulling the particle below the straight-line path. To travel in a perfectly straight line without dropping, the particle would need to travel infinitely fast! Therefore, Option D is incorrect.