Question

6.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

Question 24490

The equations of motion of bullet are: $s_y = \frac{4}{5} v_0 t - \frac{g t^2}{2}$ ; $s_x = \frac{3}{5} v_0 t$ , where $v_0$ is initial velocity to be found. The vertical coordinate of stone as a function of time is $s_y = h_0 - \frac{g t^2}{2}$ . In order to hit the stone $s_y$ of stone and bullet must be equal. This gives $\frac{4}{5} v_0 t = 50$ or $v_0 t = 50$ . Time to move to vertically maximum position is $t = \frac{4}{5} \frac{v_0}{g}$ . Hence, plugging this into latter expression, obtain $v_0 = \frac{50 \cdot 5 \cdot g}{4 v_0} \approx 63.7 \, \text{m/s}$ .