Projectile Motion Trick: Time of Flight Ratio in 60 Seconds! 🔥

 

❓ Question

Two projectiles are fired from the ground with the same initial speed from the same point, at angles

$$({45}^{\circ }+\alpha )\text{and}({45}^{\circ }-\alpha )$$

with the horizontal.

Find the ratio of their times of flight.

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🖼️ Question Image

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✍️ Short Solution

This is a classic JEE symmetry question from projectile motion.
The trick is to remember what depends on sine and what depends on sine double-angle.

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🔹 Step 1 — Time of flight formula

For a projectile fired from ground with speed $u$ at angle $\theta$:

$$T=\frac{2u\sin \theta }{g}$$

📌 Time of flight depends on $\sin\theta$ (not on $\sin 2\theta$).

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🔹 Step 2 — Write times for both projectiles

Let:

• $T_1$ = time of flight at angle $(45^\circ+\alpha)$
  

• $T_2$ = time of flight at angle $({45}^{\circ }-\alpha )$

Using the formula:

$$T_1=\frac{2u\sin(45^\circ+\alpha)}{g}$$
$$T_2=\frac{2u\sin(45^\circ-\alpha)}{g}$$

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🔹 Step 3 — Take the ratio

$$\frac{T_1}{T_2}=\frac{\sin ({45}^{\circ }+\alpha )}{\sin ({45}^{\circ }-\alpha )}$$

Constants $\frac{2u}{g}$ cancel out.

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🔹 Step 4 — Use sine angle identities

$$\sin(45^\circ+\alpha)=\frac{1}{\sqrt{2}}(\cos\alpha+\sin\alpha)$$
$$\sin(45^\circ-\alpha)=\frac{1}{\sqrt{2}}(\cos\alpha-\sin\alpha)$$

Substitute:

$$\frac{T_1}{T_2}=\frac{\cos \alpha +\sin \alpha }{\cos \alpha -\sin \alpha }$$

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🔹 Step 5 — Final ratio

So the ratio of times of flight is:

$$\boxed{{\displaystyle \frac{T_1}{T_2}=\frac{\cos \alpha +\sin \alpha }{\cos \alpha -\sin \alpha }}}$$

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✅ Final Answer

$$\boxed{{\displaystyle T_1:T_2=(\cos \alpha +\sin \alpha ):(\cos \alpha -\sin \alpha )}}$$

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🔥 Important JEE Insight (VERY EXAM-FRIENDLY)

• Angles $(45^\circ+\alpha)$ and $(45^\circ-\alpha)$ give:

  • Same range (because range $\propto \sin 2\theta$)

  • Different times of flight (because time $\propto \sin\theta$)

📌 This is why JEE loves this pair of angles:

• Range symmetry ✔️

• Time asymmetry ✔️