Two bodies X and Y are projected on the same horizontal plane, with the same initial speed but at angles 30 and 60 respectively to the horizontal. Neglecting air resistance, the ra... Two bodies X and Y are projected on the same horizontal plane, with the same initial speed but at angles 30 and 60 respectively to the horizontal. Neglecting air resistance, the ratio of the range of X to that of Y is? Read more

Steps to Solve

1. Understand the Range Formula The range ( R ) of a projectile launched at an angle ( \theta ) with initial speed ( u ) is given by the formula:
   $$ R = \frac{u^2 \sin(2\theta)}{g} $$
   where ( g ) is the acceleration due to gravity.

2. Calculate the Range for Projectile X For body X launched at ( 30^\circ ):
   $$ R_X = \frac{u^2 \sin(2 \cdot 30)}{g} $$
   Knowing ( \sin(60^\circ) = \frac{\sqrt{3}}{2} ):
   $$ R_X = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} $$

3. Calculate the Range for Projectile Y For body Y launched at ( 60^\circ ):
   $$ R_Y = \frac{u^2 \sin(2 \cdot 60)}{g} $$
   Knowing ( \sin(120^\circ) = \frac{\sqrt{3}}{2} ):
   $$ R_Y = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} $$

4. Find the Ratio of Ranges Now, we find the ratio of the ranges ( \frac{R_X}{R_Y} ):
   $$ \frac{R_X}{R_Y} = \frac{\frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g}}{\frac{u^2 \cdot 1}{g}} $$
   This simplifies to:
   $$ \frac{R_X}{R_Y} = \frac{\frac{\sqrt{3}}{2}}{1} = \frac{\sqrt{3}}{2} $$

5. Final Ratio Calculation So the final ratio of the ranges is:
   $$ \frac{R_X}{R_Y} = \frac{\sqrt{3}}{2} $$