# How to Handle the Collision of Projectiles Explained by Fully Worked Solutions

Collision of Projectile Motion is handled by setting the same vertical height and horizontal distance at a given time.

## Worked Examples of Collision of Projectile Motion

Two Points $A$ and $B$ are $d$ metres apart. A particle is projected from $A$ towards $B$ with initial velocity $u$ m/s at angle $\alpha$ to the horizontal. Concurrently, another particle is projected from $B$ towards $A$ with initial velocity $w$ m/s at $\beta$ to the horizontal. The particles collide with projectile motion when they both reach their maximum height.

The equation of projectile motion projected from the origin with initial velocity $V$ m/s at angle $\theta$ to the horizontal is $x = Vt \cos \theta$ and $\displaystyle y=Vt \sin \theta – \frac{gt^2}{2}$.

(a)    Find the time taken from $A$ to reach its maximum height.

The maximum height occurs when $\dot{y} = 0$.
\begin{aligned} \dot{y} &= u \sin \alpha-gt \\ 0 &= u \sin \alpha-gt \\ gt &= u \sin \alpha \\ \therefore t &= \frac{u \sin \alpha}{g} \end{aligned}

(b)    Show that $u \sin \alpha = w \sin \beta$.

The particle from $B$ reaches its maximum height after $\displaystyle \frac{w \sin \beta}{g}$. Both particles reach their maximum height concurrently upon collision.
\begin{aligned} \displaystyle \frac{u \sin \alpha}{g} = \frac{w \sin \beta}{g} \\ \therefore u \sin \alpha = w \sin \beta \end{aligned}

(c)    Show that $\displaystyle d = \frac{uw}{g} \sin(\alpha + \beta)$.

\begin{aligned} \displaystyle \require{color} x_A &= V t \cos \theta \\ &= u \times \frac{u \sin \alpha}{g} \times \cos \alpha \require{AMSsymbols} &\color{red} \text{horizontal distance from } A \text{ at } t = \frac{u \sin \alpha}{g} \\ &= \frac{u^2}{g} \sin \alpha \cos \alpha \\ x_B &= V t \cos \theta \\ &= w \times \frac{w \sin \beta}{g} \times \cos \beta&\color{red} \text{horizontal distance from } B \text{ at } t = \frac{w \sin \beta}{g} \\ &= \frac{w^2}{g} \sin \beta \cos \beta \\ d &= x_A + x_B \\ &= \frac{u^2}{g} \sin \alpha \cos \alpha + \frac{w^2}{g} \sin \beta \cos \beta \\ &= \frac{1}{g}(u \sin \alpha \times u \cos \alpha + w \sin \beta \times w \cos \beta) \\ &= \frac{1}{g}(w \sin \beta \times u \cos \alpha + u \sin \alpha \times w \cos \beta) &\color{red} u \sin \alpha = w \sin \beta \\ &= \frac{uw}{g}(\sin \beta \times \cos \alpha + \sin \alpha \times \cos \beta) \\ \therefore d &= \frac{uw}{g} \sin(\alpha + \beta) \end{aligned}

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