What is the range R for same-speed same-range projectiles with flight times 5s and 10s?

R=gT₁T₂/2=10×5×10/2=250m. The formula R=gT₁T₂/2 applies when two projectiles have complementary angles.

Why is R=gT₁T₂/2?

For complementary angles: T₁=2usinθ/g, T₂=2ucosθ/g. T₁T₂=4u²sinθcosθ/g²=2u²sin2θ/g²=2R/g. So R=gT₁T₂/2.

Two projectiles with same speed same range, times 5s and 10s — find range R

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PhysicsProjectile

QTwo 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/s²)

A) 250 B) 25 C) 500 D) 125Projectile Motion

MCQ

✅ Correct Answer

A) 250 m

✓

Solution

Use the property of complementary angles

Two projectiles with the same speed and same range must be launched at complementary angles ($\theta$ and $90°-\theta$).

$T_1=\dfrac{2u\sin\theta}{g}=5\text{ s}$

$T_2=\dfrac{2u\cos\theta}{g}=10\text{ s}$

(since $\sin(90°-\theta)=\cos\theta$)

Use the product formula

$T_1\cdot T_2=\dfrac{2u\sin\theta}{g}\cdot\dfrac{2u\cos\theta}{g}=\dfrac{4u^2\sin\theta\cos\theta}{g^2}=\dfrac{2u^2\sin2\theta}{g^2}$

$R=\dfrac{u^2\sin2\theta}{g}$

$\therefore\quad T_1\cdot T_2=\dfrac{2R}{g}$

$R=\dfrac{g\cdot T_1\cdot T_2}{2}=\dfrac{10\times5\times10}{2}=\boxed{250\text{ m}}$

📘 T₁·T₂ = 2R/g for Same-Range Projectiles

For two projectiles with same speed covering the same range (complementary angles $\theta$ and $90°-\theta$): $T_1=2u\sin\theta/g$, $T_2=2u\cos\theta/g$. Product: $T_1T_2=2u^2\sin2\theta/g^2=2R/g$. So $R=gT_1T_2/2$.

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Important Concepts

Complementary Angles TheoremSame speed, same range → complementary launch angles. If one angle is θ, other is 90°−θ. Their sines are complementary: sin(90°−θ)=cosθ.

Key Formula: R=gT₁T₂/2This is a very useful result: for complementary angle projectiles, R=g×T₁×T₂/2. Here: R=10×5×10/2=250m.

Time of Flight FormulaT=2u·sinθ/g. For angle θ: T₁=2u·sinθ/g. For angle (90°−θ): T₂=2u·cosθ/g. Their product gives 2u²·sin2θ/g²=2R/g.

Range FormulaR=u²·sin2θ/g. Note sin2θ=2sinθ·cosθ. Product T₁T₂=4u²sinθcosθ/g²=2u²sin2θ/g²=2R/g.

FAQs

Why do the two projectiles have complementary angles?

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Same speed + same range requires sin2θ₁=sin2θ₂. This is satisfied when θ₁+θ₂=90°, i.e., complementary angles.

Can I find u and θ separately?

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From T₁=5s: 2u·sinθ=50. From T₂=10s: 2u·cosθ=100. So tanθ=50/100=0.5, and u=50/(2sin(tan⁻¹0.5))≈55.9 m/s. But the formula R=gT₁T₂/2 avoids this.

What if I use R=u²sin2θ/g directly?

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Need u and θ. The shortcut T₁T₂=2R/g avoids finding them. Always use this product formula for JEE.

Verify: using u and θ

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sin θ=5g/(2u)=5×10/(2×55.9)≈0.447, θ≈26.6°. R=55.9²×sin(53.2°)/10≈55.9²×0.8/10≈250m ✓

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026.

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