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$$^2$$)

Solution :

For projectile motion,

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

Given time of flights :

$$T_1 = 5\text{ s}$$

$$T_2 = 10\text{ s}$$

For same horizontal range with same speed, projection angles are complementary.

Hence,

$$\sin\theta_2 = \cos\theta_1$$

Now,

$$T_1T_2=\frac{2u\sin\theta_1}{g}\times\frac{2u\cos\theta_1}{g}$$

$$=\frac{4u^2\sin\theta_1\cos\theta_1}{g^2}$$

Using,

$$2\sin\theta\cos\theta = \sin2\theta$$

$$T_1T_2=\frac{2u^2\sin2\theta_1}{g^2}$$

Range of projectile :

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

Therefore,

$$T_1T_2 = \frac{2R}{g}$$

Substituting values :

$$5 \times 10 = \frac{2R}{10}$$

$$50 = \frac{R}{5}$$

$$R = 250\text{ m}$$

Final Answer :

$$250\text{ m}$$