Physics Q83 – AP-EAMCET 2026

Home Physics Motion in a PlaneQuestion #296

Two bodies are projected from the same point at angles

15^{\circ}

and

45^{\circ}

with respect to the horizontal. If they attain the same horizontal range, the ratio of their initial velocities

Clue 1 of 3: Projectile Motion Range

The horizontal range of a projectile is determined by its initial velocity, the angle of projection, and the acceleration due to gravity.

🧠 Full Solution Path

Step 1: Recall the Horizontal Range Formula✦ Active

For a projectile launched with an initial velocity $u$ at an angle $θ$ with respect to the horizontal, the horizontal range $R$ is given by the formula:

R = \frac{u^{2} \sin (2 θ)}{g}

where $g$ is the acceleration due to gravity.

💡 Teacher's Secret Hint

Ensure you use the angle $2 θ$ in the sine function, not just $θ$ .

Step 2: Set Up Equations for Both Bodies○ Expand

Let $u_{1}$ and $θ_{1}$ be the initial velocity and angle for the first body, and $u_{2}$ and $θ_{2}$ for the second body. We are given:

$θ_{1} = 15^{\circ}$

$θ_{2} = 45^{\circ}$

The problem states that they attain the same horizontal range, so $R_{1} = R_{2}$ . Thus, we can write:

\frac{u_{1}^{2} \sin (2 θ_{1})}{g} = \frac{u_{2}^{2} \sin (2 θ_{2})}{g}

💡 Teacher's Secret Hint

Always clearly define your variables before starting calculations to avoid confusion.

Step 3: Substitute Values and Simplify○ Expand

We can cancel $g$ from both sides and substitute the given angles:

u_{1}^{2} \sin (2 \times 15^{\circ}) = u_{2}^{2} \sin (2 \times 45^{\circ})

This simplifies to:

u_{1}^{2} \sin (30^{\circ}) = u_{2}^{2} \sin (90^{\circ})

Recall the values for $\sin (30^{\circ}) = \frac{1}{2}$ and $\sin (90^{\circ}) = 1$ . Substitute these values:

u_{1}^{2} (\frac{1}{2}) = u_{2}^{2} (1)

💡 Teacher's Secret Hint

Remember common trigonometric values. If you forget, you can quickly sketch a $30 - 60 - 90$ or $45 - 45 - 90$ triangle.

Step 4: Calculate the Ratio of Initial Velocities○ Expand

Rearrange the equation to find the ratio $\frac{u_{1}}{u_{2}}$ :

\frac{u_{1}^{2}}{u_{2}^{2}} = \frac{1}{1 / 2} = 2

Taking the square root of both sides:

\frac{u_{1}}{u_{2}} = \sqrt{2}

Therefore, the ratio of their initial velocities is $\sqrt{2} : 1$ .

💡 Teacher's Secret Hint

Always ensure you're solving for the correct ratio (e.g., $u_{1} : u_{2}$ vs $u_{2} : u_{1}$ ) as requested in the question.

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Physics Question 83 – AP-EAMCET 2026

Clue 1 of 3: Projectile Motion Range

🧠 Full Solution Path