Question

body projected at such an angle that horizontal range is 3 times greatest height.find angle of projection

Answer 1

Answer on Question #43993 – Physics - Mechanics | Kinematics | Dynamics

body projected at such an angle that horizontal range is 3 times greatest height.find angle of projection

Solution:

Equation of the motion for body, thrown at angle $\alpha$ : (L - maximum horizontal range of this body) (t - time of the flight)

$$u _ {x} = u \cos \alpha ; u _ {y} = u \sin \alpha ;$$ $$x: L = u t \cos \alpha$$ $$y: 0 = u t \sin \alpha - \frac {g t ^ {2}}{2}$$ $$u \sin \alpha = \frac {g t}{2}$$ $$t = \frac {2 u \sin \alpha}{g}$$

(2) in (1):

$$L = u \frac {2 u \sin \alpha}{g} \cos \alpha = \frac {2 u ^ {2} \sin \alpha \cos \alpha}{g}$$

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

$$t _ {1} = \frac {t}{2} = \frac {u \sin \alpha}{g}$$ $$y: (\text {half of the flight}): h _ {1} = u t _ {1} \sin \alpha - \frac {g t _ {1} ^ {2}}{2}$$ $$h = u t _ {1} \sin \alpha - \frac {g t _ {1} ^ {2}}{2}$$ $$h = u \frac {u \sin \alpha}{g} \sin \alpha - \frac {g \left(\frac {u \sin \alpha}{g}\right) ^ {2}}{2} = \frac {u ^ {2} \sin^ {2} \alpha}{2 g}$$

From the problem statement:

$$L = 3 \cdot h$$

(4) and (3) in (5):

$$\frac {2 u ^ {2} \sin \alpha \cos \alpha}{g} = \frac {3 \cdot u ^ {2} \sin^ {2} \alpha}{2 g}$$ $$4 \cos \alpha = 3 \sin \alpha$$ $$\frac {4}{3} = \tan \alpha \Rightarrow \alpha = \arctan \left(\frac {4}{3}\right) = 53{}^{\circ}$$

Answer: angle of projection if equal to $53{}^{\circ}$.

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