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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