Question

Two particles projected from the same point with equal speeds at an angle α &β strikes the horizontal ground at the same point. If H1 & H2 be maximum height be the range for both, T1 & T2 is the time of flight. Then
1. α +β=½
2. R=4H1H2
3. T1/T2=tanα
4. Tanα=h1/h2

Answer 1

Answer on Question#38191 - Physics - Mechanics

Two particles projected from the same point with equal speeds at an angle $\alpha$ & $\beta$ strikes the horizontal ground at the same point. If H1 & H2 be maximum height be the range for both, T1 & T2 is the time of flight. Then

1. $\alpha +\beta = 1 / 2$

2. $R = 4H1H2$

3. T1/T2=tanα

4. Tanα=h1/h2

Solution:

We can write the equations of motion for each of the bodies:

Equations for body, thrown at angle $\alpha$ : (L1 - maximum range of this body)

$$V _ {x} = V \cos \alpha ; V _ {y} = V \sin \alpha ;$$ $$x: L _ {1} = V T _ {1} \cos \alpha (1), T _ {1} - \text {time of the flight}$$ $$y: 0 = V T _ {1} \sin \alpha - \frac {g T _ {1} ^ {2}}{2}$$ $$V \sin \alpha = \frac {g T _ {1}}{2}$$ $$T _ {1} = \frac {2 V \sin \alpha}{g} (2)$$ $$(2) \text {in} (1): L _ {1} = V \frac {2 V \sin \alpha}{g} \cos \alpha = \frac {2 V ^ {2} \sin \alpha \cos \alpha}{g}$$

Maximum height: the time taken to reach the maximum height is equal to half of the time of flight:

$$t _ {1} = \frac {T _ {1}}{2} = \frac {V \sin \alpha}{g}$$

y: (half of the flight): $H_1 = Vt_1 \sin \alpha - \frac{gt_1^2}{2}$

$$H_1 = Vt_1 \sin \alpha - \frac{gt_1^2}{2}$$ $$H_1 = V \frac{V \sin \alpha}{g} \sin \alpha - \frac{g \left(\frac{V \sin \alpha}{g}\right)^2}{2} = \frac{V^2 \sin^2 \alpha}{2g}$$

Equation for body, thrown at angle $\beta$ (bodies strikes the horizontal ground at the same point):

$$L_2 = V \frac{2V \sin \beta}{g} \cos \beta = \frac{2V^2 \sin \beta \cos \beta}{g} = L_1$$ $$\frac{2V^2 \sin \alpha \cos \alpha}{g} = \frac{2V^2 \sin \beta \cos \beta}{g}$$ $$\sin 2\alpha = \sin 2\beta$$ $$2\alpha = \pi - 2\beta$$ $$2\alpha + 2\beta = \pi$$ $$\alpha + \beta = \frac{1}{2} \pi$$

Answer: 1) $\alpha + \beta = \frac{1}{2} \pi$.