Question

the horizontal distance covered by a projectile in motion is maximized at which angle?

Answer 1

The horizontal distance covered by a projectile in motion is maximized at which angle?

Solution.

$$v _ {x} = v _ {0} \cos \alpha ;$$ $$v _ {y} = v _ {0} \sin \alpha .$$

The max distance.

$$l = v _ {0 x} t;$$ $$l = v _ {0} \cos \alpha t;$$

The time of flight.

$$y = v _ {0 y} t - \frac {g t ^ {2}}{2};$$ $$y = v _ {0} \sin \alpha t - \frac {g t ^ {2}}{2}.$$

At the end of the flight $y = 0$ :

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

The max distance.

$$l = v _ {0} \cos \alpha \frac {2 v _ {0} \sin \alpha}{g};$$ $$l = \frac {2 v _ {0} ^ {2} \sin \alpha \cos \alpha}{g}.$$

When the distance is maximized - the derivative of a function of the distance is zero (an angle is the independent variable):

$$l _ {\max } ^ {\prime} = 0;$$ $$\left(\frac {2 v _ {0} ^ {2} \sin \alpha \cos \alpha}{g}\right) ^ {\prime} = \frac {2 v _ {0} ^ {2}}{g} \left((\sin \alpha) ^ {\prime} \cos \alpha - \sin \alpha (\cos \alpha) ^ {\prime}\right) =$$ $$= \frac {2 v _ {0} ^ {2}}{g} (\cos \alpha \cos \alpha - \sin \alpha \sin \alpha) = \frac {2 v _ {0} ^ {2}}{g} (\cos^ {2} \alpha - \sin^ {2} \alpha);$$ $$\frac {2 v _ {0} ^ {2}}{g} \left(\cos^ {2} \alpha - \sin^ {2} \alpha\right) = 0;$$ $$\cos^ {2} \alpha = \sin^ {2} \alpha ;$$ $$\tan^ {2} \alpha = 1;$$ $$\tan \alpha = 1;$$ $$\arctan 1 = 45 {}^ {\circ};$$ $$\alpha = 45 {}^ {\circ}.$$

**Answer:**

The horizontal distance covered by a projectile in motion is maximized at $\alpha = 45{}^{\circ}$.