Question

Two particles are projected from a tower of height 400 m & angles 45° & 60° horizontally. If they have the same time of flight, find the ratio of their velocities.

• A. \(\frac{\sqrt{6}}{2}\)
• B. \(\frac{\sqrt5}{\sqrt2}\)
• C. \(\frac{\sqrt3}{\sqrt4}\)
• D. 1

Answer 1

The Correct Option is A

The problem asks for the ratio of the initial velocities of two particles projected from the top of a 400 m tower at different angles, given that they have the same time of flight.

Concept Used:

The solution is based on the analysis of the vertical component of projectile motion. The time of flight of a projectile launched from a height depends on the initial vertical velocity, the height of projection, and the acceleration due to gravity.

The equation for the vertical displacement (\(\Delta y\)) of a projectile is:

\[ \Delta y = v_{y0} t + \frac{1}{2} a_y t^2 \]

where \(v_{y0}\) is the initial vertical velocity, \(a_y\) is the vertical acceleration (which is \(-g\)), and \(t\) is the time.

For a projectile launched at an angle \(\theta\) with the horizontal with an initial velocity \(v\), the initial vertical velocity is \(v_{y0} = v \sin \theta\).

Step-by-Step Solution:

Step 1: Establish a coordinate system and define the parameters.

Let's set the origin \((0, 0)\) at the point of projection on top of the tower. The downward direction is taken as negative.

• Initial height: \(y_0 = 0\)
• Final height (ground): \(y = -h = -400\) m
• Vertical acceleration: \(a_y = -g\)
• Time of flight: \(t = T\)

The angles of projection are given as \(\theta_1 = 45^\circ\) and \(\theta_2 = 60^\circ\) with the horizontal.

Step 2: Write the equation for the time of flight.

Using the equation for vertical displacement, when the particle hits the ground at time \(t = T\), its vertical displacement is \(\Delta y = -h\). The initial vertical velocity is \(v_{y0} = v \sin \theta\).

\[ -h = (v \sin \theta) T - \frac{1}{2}gT^2 \]

Step 3: Relate the initial vertical velocity to the time of flight.

We can rearrange the equation from Step 2 to solve for the initial vertical velocity term, \(v \sin \theta\).

\[ (v \sin \theta) T = \frac{1}{2}gT^2 - h \] \[ v \sin \theta = \frac{gT}{2} - \frac{h}{T} \]

This equation shows that for a fixed height \(h\) and a specific time of flight \(T\), the initial vertical velocity \(v_{y0} = v \sin \theta\) is uniquely determined.

Step 4: Apply the condition that the time of flight is the same for both particles.

The problem states that both particles have the same time of flight, i.e., \(T_1 = T_2 = T\). Since \(g\) and \(h\) are also the same for both, it follows from the equation in Step 3 that their initial vertical velocities must be equal.

\[ v_{y1} = v_{y2} \] \[ v_1 \sin \theta_1 = v_2 \sin \theta_2 \]

Note that the height of the tower \(h = 400\) m is not needed to find the ratio, as the condition of equal times of flight directly implies equal initial vertical velocities.

Step 5: Calculate the ratio of the initial velocities.

From the equality derived in Step 4, we can find the ratio \(v_1 / v_2\).

\[ \frac{v_1}{v_2} = \frac{\sin \theta_2}{\sin \theta_1} \]

Substitute the given angles \(\theta_1 = 45^\circ\) and \(\theta_2 = 60^\circ\).

\[ \frac{v_1}{v_2} = \frac{\sin 60^\circ}{\sin 45^\circ} \]

Final Computation & Result:

We use the standard values for the sine functions:

• \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)
• \( \sin 45^\circ = \frac{1}{\sqrt{2}} \)

Substituting these values into the ratio:

\[ \frac{v_1}{v_2} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}} = \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{1} = \frac{\sqrt{6}}{2} \]

The ratio of their velocities is \( \frac{\sqrt{6}}{2} \).