Question

Your dorm wishes to purchase a small used cannon with the intent of firing a blank after each touchdown at home games. The spec sheet for a cannon of interest is missing some pages; however, the information you have claims that when the cannon is fired, the position of the cannonball as a function of time (while it is in the barrel of the cannon) is given by

x(t) = (1.50 ✕ 10⁵ m/s²)t² − (1.20 ✕ 10⁷ m/s³)t³.

The spec sheet further claims that the cannon design is such that the acceleration of the cannonball is zero at the exit end of the barrel. In order to learn more about the cannon, you are asked to determine the following.

(a) time it takes for a cannonball to travel the length of the barrel
s

(b) speed of a cannonball as it exits the barrel
m/s

(c) length of the barrel
m

Answer 1

Given,

x(t) = 1.5 x 10^5 m/s^2 t^2 - 1.2 x 10^7 m/s^3 t^3

a)for the time, x(t) = 0

1.5 x 10^5 m/s^2 t^2 = 1.2 x 10^7 m/s^3 t^3

t = 1.5 x 10^5/(1.2 x 10^7) = 0.0125 m

Hence, t = 0.0125 m

b)We know that

v = dx/dt

v = d[1.5 x 10^5 m/s^2 t^2 - 1.2 x 10^7 m/s^3 t^3]/dt

v = 3 x 10^5 t - 3.6 x 10^7 t^2

putting t = 0.0125

v = 3 x 10^5 x 0.0125 - 3.6 x 10^7 x 0.0125^2 = 1875 m/s

Hence, v = 1875 m/s

c)v = l/t => l = v t

l = 1875 x 0.0125 = 23.44 m

l = 23.44 m