Motion in a Plane

From graph,

${v_{RG}} = 15\sqrt 2 \tan 45^\circ $

$ = 15\sqrt 2 $

$ = {{30} \over {\sqrt 2 }}$

When two projectiles are thrown with the same initial velocity 'u' but at complementary angles (say, $\theta$ and $(90^\circ - \theta)$) with the horizontal, they attain the same range R. The formula for the range R of a projectile is:

$ R = \frac{u^2 \sin 2\theta}{g} $

For complementary angles, $2 \theta$ and $180^\circ - 2 \theta$ (which simplifies to the same value for the sine function), the ranges are equal.

The maximum height $h_1$ for angle $\theta$ is given by:

$ h_1 = \frac{u^2 \sin^2 \theta}{2g} $

And the maximum height $h_2$ for angle $(90^\circ - \theta)$ is given by:

$ h_2 = \frac{u^2 \cos^2 \theta}{2g} $

Now, multiplying these heights:

$ h_1 h_2 = \left( \frac{u^2 \sin^2 \theta}{2g} \right) \cdot \left( \frac{u^2 \cos^2 \theta}{2g} \right)$

This simplifies to:

$ h_1 h_2 = \frac{u^4 \sin^2 \theta \cos^2 \theta}{4g^2}$

Since $\sin^2 \theta \cos^2 \theta = \left( \frac{\sin 2 \theta}{2} \right)^2 = \frac{1}{4} \sin^2 2 \theta$:

$ h_1 h_2 = \frac{u^4 \sin^2 2 \theta}{16g^2}$

Using the range formula $ R = \frac{u^2 \sin 2 \theta}{g} $, we get:

$ R^2 = \left( \frac{u^2 \sin 2 \theta}{g} \right)^2$

Thus, we have:

$ 4h_1 h_2 = \frac{u^4 \sin^2 2 \theta}{4g^2} = \frac{R^2}{4} $

This simplifies to:

$ R = 4\sqrt{h_1 h_2}$

Both the assertion and the reason are correct, and the reason correctly explains the assertion.

The correct answer is:

Option A : Both A and R are true and R is the correct explanation of A.

$t = {{25\sin \theta } \over g}$

and, $R = {{{{(25)}^2}(2\sin \theta \cos \theta )} \over g}$

$ \Rightarrow R = {{25 \times 25 \times 2} \over g} \times {{gt} \over {25}} \times \cos \theta $

$ \Rightarrow R = 50t\cos \theta $

$\therefore$ $tan\theta = {{gt} \over {25}} \times {{50t} \over R}$

$ = {{20{t^2}} \over R}$

$ \Rightarrow \theta = {\cot ^{ - 1}}\left( {{R \over {20{t^2}}}} \right)$

The correct answer is Option C, a straight line vertically down the plane.

Here's why:

• Frame of Reference: The key to understanding projectile motion is considering the frame of reference. The observer in the plane is in a moving frame of reference. From their perspective, they are at rest, and the bomb has the same horizontal velocity as the plane.


• Gravity's Influence: The only force acting on the bomb after it's released is gravity. Gravity acts vertically downwards.


• Resultant Motion: Because the bomb has the same horizontal velocity as the plane and is only affected by gravity vertically, it appears to the observer in the plane to fall straight down.

Let's eliminate the other options:

• Option A (Hyperbola): A hyperbolic trajectory occurs when an object is influenced by two forces acting in different directions, resulting in a path that gets progressively straighter. This doesn't apply to the bomb scenario.


• Option B and D (Parabola): A parabolic trajectory is observed from a stationary frame of reference on the ground. In this case, the horizontal motion of the bomb combined with the vertical acceleration due to gravity creates a parabolic path.

Considering the initial position of ship A as origin, so the velocity and position of ship will be

${\overrightarrow v _A} = (30\widehat i + 50\widehat j)$ and ${\overrightarrow r _A} = (0\widehat i + 0\widehat j)$

Now, as given in the question, velocity and position of ship B will be, ${\overrightarrow v _B} = - 10\widehat i$ and ${\overrightarrow r _B} = (80\widehat i + 150\widehat j)$

Time after which the distance is minimum between A and B can be calculated as

$t = {{|{{\overrightarrow r }_{BA}}.\,{{\overrightarrow v }_{BA}}|} \over {|{{\overrightarrow v }_{BA}}{|^2}}}$

where, ${\overrightarrow r _{BA}} = {\overrightarrow r _B} - {\overrightarrow r _A} = 80\widehat i + 150\widehat j$

and ${\overrightarrow v _{BA}} = - 10\widehat i - (30\widehat i + 50\widehat j)$

$ = - 40\widehat i - 50\widehat j$

$ \Rightarrow t = {{|(80\widehat i + 150\widehat j)\,.\,( - 40\widehat i - 50\widehat j)|} \over {| - 40\widehat i - 50\widehat j{|^2}}}$

$ = {{3200 + 7500} \over {4100}} = {{10700} \over {4100}} = 2.6$ h

Let the distance QM = l and distance RM = x.

Time to reach from Q to R is ${t_1} = {{l - x} \over v}$

Time to reach from R to P is ${t_2} = {{\sqrt {{x^2} + {d^2}} } \over {v/2}}$

Therefore, $t = {t_1} + {t_2} = {{l - x} \over v} + {{\sqrt {{x^2} + {d^2}} } \over {v/2}}$

On differentiating, we get

${{dt} \over {dx}} = {{0 - 1} \over v} + {1 \over {v/2}}{1 \over {2\sqrt {{x^2} + {d^2}} }} \times 2x$

$ \Rightarrow {{dt} \over {dx}} = {{ - 1} \over v} + {{2x} \over {v\sqrt {{x^2} + {d^2}} }}$

Put ${{dt} \over {dx}} = 0$, we get

${{ - 1} \over v} + {{2x} \over {v\sqrt {{x^2} + {d^2}} }} = 0$

$ \Rightarrow {{2x} \over {v\sqrt {{x^2} + {d^2}} }} = {1 \over v}$

$ \Rightarrow 2x = \sqrt {{x^2} + {d^2}} \Rightarrow 4{x^2} = {x^2} + {d^2}$

$ \Rightarrow 3{x^2} = {d^2} \Rightarrow x = {d \over {\sqrt 3 }}$

Therefore, the distance $RM = x = {d \over {\sqrt 3 }}$, the time taken to reach P is minimum.