Collision of Two Projectiles
Question
As shown in figure, a body is projected with velocity \( u_1 \) from the point O at an angle of 60° to the horizontal. At the same time, another body is projected vertically upwards with velocity \( u_2 \) from the point B, which is at a horizontal distance from O. Find the value of \( \frac{u_1}{u_2} \) so that both the bodies collide.Answer
Let the two bodies collide at time \( t \) after projection.
For the body projected from O:
- Initial velocity = \( u_1 \)
- Angle of projection = 60°
- Horizontal component of velocity = \( u_1 \cos 60^\circ = \frac{u_1}{2} \)
- Vertical component of velocity = \( u_1 \sin 60^\circ = \frac{\sqrt{3}}{2} u_1 \)
Horizontal position at time \( t \):
x = u_1 \cos 60^\circ \cdot t = \frac{u_1}{2} t
Vertical position at time \( t \):
y = u_1 \sin 60^\circ \cdot t - \frac{1}{2} g t^2 = \frac{\sqrt{3}}{2} u_1 t - \frac{1}{2} g t^2
For the body projected vertically upwards from B:
- Initial velocity = \( u_2 \)
- Horizontal position = constant = distance OB = \( x_B \)
- Vertical position at time \( t \):
y = u_2 t - \frac{1}{2} g t^2
Since the bodies collide, their positions must be equal at time \( t \):
- Horizontal positions equal:
- Vertical positions equal:
x = x_B = \frac{u_1}{2} t
\frac{\sqrt{3}}{2} u_1 t - \frac{1}{2} g t^2 = u_2 t - \frac{1}{2} g t^2
Simplify vertical position equality:
\frac{\sqrt{3}}{2} u_1 t = u_2 t
\Rightarrow \frac{\sqrt{3}}{2} u_1 = u_2
Rearranging for \( \frac{u_1}{u_2} \):
\frac{u_1}{u_2} = \frac{2}{\sqrt{3}}
Final Answer
\boxed{\frac{u_1}{u_2} = \frac{2}{\sqrt{3}}}