Physics Q31 – JEE-MAIN 2026

Home Physics Motion in a PlaneQuestion #306

Two identical bodies, projected with the same speed at two different angles cover the same horizontal range R. If the time of flight of these bodies are

5 s

and

10 s

, respectively, then the value of R is _______ m. (Take

g = 10 {m/s}^{2}

)

Clue 1 of 3: Complementary Angles

When two projectiles launched with the same initial speed achieve the same horizontal range, their projection angles must be complementary.

🧠 Full Solution Path

Step 1: Identify complementary angles and express time of flights✦ Active

Since the horizontal range $R$ is the same for two different angles with the same initial speed $u$ , the angles must be complementary. Let the angles be $θ$ and $90^{\circ} - θ$ .

The time of flight for angle $θ$ is $t_{1} = \frac{2 u \sin θ}{g} = 5 s$ .

The time of flight for angle $90^{\circ} - θ$ is $t_{2} = \frac{2 u \sin (90^{\circ} - θ)}{g} = \frac{2 u \cos θ}{g} = 10 s$ .

Step 2: Derive expressions for

u \sin θ

and

u \cos θ

○ Expand

From the time of flight equations, we can write:

u \sin θ = \frac{5 g}{2}

u \cos θ = \frac{10 g}{2} = 5 g

Step 3: Calculate the horizontal range

R

○ Expand

The horizontal range is given by $R = \frac{u^{2} \sin (2 θ)}{g}$ . Using the identity $\sin (2 θ) = 2 \sin θ \cos θ$ , we get:

R = \frac{u^{2} (2 \sin θ \cos θ)}{g} = \frac{2}{g} (u \sin θ) (u \cos θ)

Substitute the expressions for $u \sin θ$ and $u \cos θ$ from Step 2:

R = \frac{2}{g} (\frac{5 g}{2}) (5 g) = \frac{2}{g} (\frac{25 g^{2}}{2}) = 25 g

Given $g = 10 {m/s}^{2}$ :

R = 25 \times 10 = 250 m

💡 Teacher's Secret Hint

Remember that $\sin (2 θ)$ can be expanded as $2 \sin θ \cos θ$ to utilize the derived terms.

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Physics Question 31 – JEE-MAIN 2026

Clue 1 of 3: Complementary Angles

🧠 Full Solution Path