Question

A stone is dropped vertically from the top of a tower of height 40 m. At
the same time a gun is aimed directly at the stone from the ground at a
horizontal distance 30 m from the base of the tower and fired. If the
bullet from the gun is to hit the stone before it reaches the ground, the
minimum velocity of the bullet must be, approximately..?

Answer 1

A stone is dropped vertically from the top of a tower of height $40\mathrm{m}$ . At the same time a gun is aimed directly at the stone from the ground at a horizontal distance $30\mathrm{m}$ from the base of the tower and fired. If the bullet from the gun is to hit the stone before it reaches the ground, the minimum velocity of the bullet must be, approximately?

Coordinate of the stone equals:

$$y _ {s} = 4 0 - \frac {g t ^ {2}}{2}$$

Coordinates of the bullet equals:

$$x _ {b} = 3 0 - v \cos \alpha t$$ $$y _ {b} = v \sin \alpha t - \frac {g t ^ {2}}{2}$$

Bullet can hit stone only in point $x = 0$ :

$$0 = 3 0 - v \cos \alpha t$$ $$t _ {0} = \frac {3 0}{v \cos \alpha}$$

And at this time $y_{s} = y_{b}$

$$4 0 - \frac {g t _ {0} ^ {2}}{2} = v \sin \alpha t _ {0} - \frac {g t _ {0} ^ {2}}{2}$$ $$40 = v \sin \alpha t_0$$ $$40 = v \sin \alpha \frac{30}{v \cos \alpha}$$ $$40 = \sin \alpha \frac{30}{\cos \alpha} = \frac{40}{50} \frac{30}{30/50} = 40$$

So, bullet hits the stone anyway, therefore, minimal velocity will be if they meet at point $(0,0)$:

$$v \sin \alpha t - \frac{g t_0^2}{2} = 0$$ $$v \sin \alpha - \frac{g \frac{30}{v \cos \alpha}}{2} = 0$$ $$v \sin \alpha = g \frac{30}{2 v \cos \alpha}$$

So, minimal velocity equals:

$$v = \sqrt{\frac{30 g}{\sin 2 \alpha}} = 25 \frac{m}{s}$$

Answer: $25 \frac{m}{s}$