Question

A small dingy is floating up to a pirate’s ship full of would be boarders. He angles his cannon downward at -15 degrees while 3.0m above the water. He fires his cannon at 95 m/s. The small boat of unfortunate crewmen is 20 m away. Do they survive their foolhardy errand?(please show work)

Answer 1

Speed of the cannon is 95 m/s

now we will find its two components

speed in horizontal direction is

$v_x = 95 cos15$

$v_x = 91.76 m/s$

speed in vertical direction is

$v_y = 95 sin(-15)$

$v_y = -24.6 m/s$

now the time taken by the cannon to reach the water surface will be find out by using kinematics

$\Delta y = v_y * t + \frac{1}{2} at^2$

$-3 = -24.6 * t - \frac{1}{2}9.8 * t^2$

$4.9t^2 + 24.6 t - 3 = 0$

By solving above quadratic equation

$t = 0.12 s$

now in the same time the horizontal distance moved by the cannon

$d_x = v_x * t$

$d_x = 91.76 * 0.12$

$d_x = 10.93 m$

Since the distance of crewmen is 20 m so it will definitely safe as cannon will hit at distance 10.93 m