Q112 / 150 · MHT CET 2026 (16 April Shift 2) · Physics
Two objects are projected with same velocity 'u' at different angles $\alpha$ and $\beta$ with the horizontal. If $\alpha + \beta = 90^\circ$, the ratio of horizontal range of the first object to the second object will be $[\sin(\pi - \theta) = \sin\theta]$
- A$4 : 1$
- B$2 : 1$
- Correct answerC$1 : 1$
- D$1 : 2$
Solution
The horizontal range of a projectile is given by $R = \dfrac{u^2 \sin(2\theta)}{g}$.
For the first object projected at angle $\alpha$, the range is $R_1 = \dfrac{u^2 \sin(2\alpha)}{g}$.
For the second object projected at angle $\beta$, the range is $R_2 = \dfrac{u^2 \sin(2\beta)}{g}$.
Given $\alpha + \beta = 90^\circ$, we have $\beta = 90^\circ - \alpha$.
Substituting $\beta$ in the expression for $R_2$:
$R_2 = \dfrac{u^2 \sin(2(90^\circ - \alpha))}{g} = \dfrac{u^2 \sin(180^\circ - 2\alpha)}{g}$
Using the identity $\sin(180^\circ - \theta) = \sin\theta$, we get:
$R_2 = \dfrac{u^2 \sin(2\alpha)}{g}$
Thus, $R_1 = R_2$, which means the ratio of their horizontal ranges is $1 : 1$.
Answer: $1 : 1$
For the first object projected at angle $\alpha$, the range is $R_1 = \dfrac{u^2 \sin(2\alpha)}{g}$.
For the second object projected at angle $\beta$, the range is $R_2 = \dfrac{u^2 \sin(2\beta)}{g}$.
Given $\alpha + \beta = 90^\circ$, we have $\beta = 90^\circ - \alpha$.
Substituting $\beta$ in the expression for $R_2$:
$R_2 = \dfrac{u^2 \sin(2(90^\circ - \alpha))}{g} = \dfrac{u^2 \sin(180^\circ - 2\alpha)}{g}$
Using the identity $\sin(180^\circ - \theta) = \sin\theta$, we get:
$R_2 = \dfrac{u^2 \sin(2\alpha)}{g}$
Thus, $R_1 = R_2$, which means the ratio of their horizontal ranges is $1 : 1$.
Answer: $1 : 1$
Hence, the correct answer is option: 'C'.