Question

A cannon fires a cannonball with muzzle velocity vo (muzzle velocity is the velocity at which the projectile leaves the cannon). The maximum range of the cannon is 17.6 km (a) What is the muzzle velocity v? (b) A target is a horizontal distance of 5.50 km away fronm the cannon-The cannon can hit this target by firing at two different angles, a low angle 6, and a high angle 0. Calculate these two angles. (c) Find the times, t and t, that the projectile takes to hit the target for both angles

Answer 1

R_mon = u^2/g rightdoublearrow 17.6 times 10^3 = v_0^2/9.81 v_0 = 415.52 m/s b) R = u^2 sin 2 theta/g 5.5 times 10^3 = 172656 times sin 2 theta/9.81 sin 2 theta = 0.1325 2 theta = sin^-1 (0.3125) = 18.21 degree theta_L = 9.11 degree degrees theta = 8.89 degree degrees c) t_L = 2 times 415.52 times sin (9.11)/9.81 = 13.41 sec t_L = 13.41 seconds t_H = 2 times 415.52 sin (80.89)89/9.81 = 83.65 sec t_H = 83.65 seconds