Question

When a particle is projected at some angles to the horizontal,it has a range R and time of flight t1. If the same particle is projected with same speed at some other angle to have the same range its time of flight is t2, then,
1. t1 + t2 = 2R/g
2. t1 - t2 = R/g
3. t1t2 = 2R/g
4. t1t2 = R/g

Answer 1

Answer on Question #48601 - Physics - Mechanics | Kinematics | Dynamics

1. When a particle is projected at some angles to the horizontal, it has a range R and time of flight t1. If the same particle is projected with same speed at some other angle to have the same range its time of flight is t2, then,

$$1. t 1 + t 2 = 2 R / g; \quad 2. t 1 - t 2 = R / g; \quad 3. t 1 t 2 = 2 R / g; \quad 4. t 1 t 2 = R / g$$$$\left\{ \begin{array}{l} x = v _ {0} \cos \alpha \cdot t \\ y = v _ {0} \sin \alpha \cdot t - \frac {g t ^ {2}}{2} \end{array} \right.$$

Here $X$-axis is directed towards the motion and $Y$-axis is directed up.

The time of falling can be found from the condition $y = 0$:

$$v _ {0} \sin \alpha \cdot t - \frac {g t ^ {2}}{2} = 0, \quad t _ {0} = \frac {2 v _ {0}}{g} \sin \alpha .$$

So, the range is

$$R = x \left(t _ {0}\right) = v _ {0} \cos \alpha \cdot t _ {0} = v _ {0} \sqrt {1 - \sin^ {2} \alpha} \cdot t _ {0} = v _ {0} \sqrt {1 - \left(\frac {g t _ {0}}{2 v _ {0}}\right) ^ {2}} \cdot t _ {0} = \frac {t _ {0}}{2} \sqrt {4 v _ {0} ^ {2} - g ^ {2} t _ {0} ^ {2}}.$$

Thus, $R^2 = \frac{t_0^2}{4} \left(4v_0^2 - g^2 t_0^2\right)$, $v_0^2 = \frac{R^2}{t_0^2} + \frac{g^2 t_0^2}{4}$.

One can write that $v_0^2 = \frac{R^2}{t_1^2} + \frac{g^2 t_1^2}{4} = \frac{R^2}{t_2^2} + \frac{g^2 t_2^2}{4}$.

Let transform the last equation.

$$\frac {R ^ {2}}{t _ {1} ^ {2}} - \frac {R ^ {2}}{t _ {2} ^ {2}} = \frac {g ^ {2} t _ {2} ^ {2}}{4} - \frac {g ^ {2} t _ {1} ^ {2}}{4}, \quad \frac {R ^ {2} \left(t _ {2} ^ {2} - t _ {1} ^ {2}\right)}{t _ {1} ^ {2} t _ {2} ^ {2}} = \frac {g ^ {2} \left(t _ {2} ^ {2} - t _ {1} ^ {2}\right)}{4}, \quad t _ {1} t _ {2} = \frac {2 R}{g}.$$

Answer: 3.

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