physics

Optimal angle for a projectile

Optimal angle for a projectile Part1

a projectile $$s:speed\quad magnitude \quad of \quad \vec{s}$$ $$\theta:angle\quad of\quad direction$$ $$d:distance(landing \quad point)$$
vertical speed $$sin\theta=\frac{s_{v}}{s} \rightarrow s_{v}=s \cdot sin\theta$$ horisontal speed $$cos\theta=\frac{s_{h}}{s} \rightarrow s_{h}=s \cdot cos\theta$$

Optimal angle for a projectile Part2

Hangtime

How long is this object going to be in the air?

$$Gravity \quad slows \quad it \quad down \quad at \quad 10m/s^2.$$ $$g=\vec{a}=-10m/s^2$$ $$\vec{s}_{v}=10m/s \rightarrow 0m/s \quad by \vec{a}:1second$$ $$\vec{s}_{v}=20m/s \rightarrow 0m/s \quad by \vec{a}:2seconds$$ $$time \quad downward =\frac{\vec{s}_{v}}{g} =\frac{20m/s}{10m/s^2}=2s$$ $$time \quad upward=time \quad downward$$ total time in the air $$t_{a}=\frac{\vec{s}_{v}}{g} \times 2 = \frac{2 \cdot s \cdot sin\theta}{g}$$

Optimal angle for a projectile Part3

Horisontal distance as a function of angle (and speed)

Horisontal distance $$d = rate \cdot time$$

$$d=r \cdot t$$ $$d=s_{h} \cdot t =s\cdot cos\theta \cdot \frac{2\cdot s \cdot sin\theta}{g}$$

a function of angle $$d(\theta)=\frac{2s^2}{g}cos\theta sin\theta \quad (0 \ge \theta \ge 90)$$


Optimal angle for a projectile Part4

Finding the optimal angle and distance with a bit of calculus

$$d(\theta)=\frac{2s^2}{g}cos\theta sin\theta \quad (0 \ge \theta \ge 90)$$ $$d(\theta)の導関数d'(\theta)を求めます$$ $$d'(\theta)=\frac{2s^2}{g}(-sin^2\theta+cos^2\theta)$$ $$導関数d'(\theta)が表すd(\theta)の接線の傾きが0のとき、d(\theta)の値は最大となります$$ $$\frac{2s^2}{g}(-sin^2\theta+cos^2\theta)=0$$ $$s=0のときは考えません$$ $$-sin^2\theta+cos^2\theta=0$$ $$cos^2\theta=sin^2\theta$$ $$\frac{cos^2\theta}{cos^2\theta}=\frac{sin^2\theta}{cos^2\theta}$$ $$1=tan^2\theta$$ $$\sqrt{1}=\sqrt{tan^2\theta}$$ $$1=tan\theta$$ $$arctan1=45°$$

The optimal angle is 45 degrees.

The optimal distance is

$$sin45°=\frac{\sqrt{2}}{2} \qquad cos45°=\frac{\sqrt{2}}{2}$$ $$d(\theta)=\frac{2s^2}{g}cos\theta sin\theta$$ $$\quad=\frac{2s^2}{g} \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}$$

$$\quad=\frac{s^2}{g}$$

$$s=10m/s \quad g=10m/s^2$$ $$d=100/10=10m$$