Question

A battleship fires two artillery shells at nearby enemy ships from the same gun, as shown in the picture below. Both artillery shells follow a symmetric trajectory, such that air resistance can be ignored, each with an initial velocity magnitude of V₀ . The ship’s deck, upon which the guns are located, is h₁ meters above sea level.

(a) The battleship’s gun initially targets ship B . Upon being fired, the artillery shell barely clears the brown mountain, at its apex, which has a height above sea-level of h₂ meters. Determine the angle of launch required for this scenario.

(b) Due to its location, when the gun is used to target ship A the sailors fire the gun and listen for the resulting explosion to check its accuracy. The sailors measure that the artillery shell was airborne for a time t before they hear the explosion from the enemy’s ship. Determine the angle of launch and the range of the artillery shell for this scenario.

(d) Calculate the numerical answers for parts (a) and (b) using the following values: v_{0=155 m/s h1=95m, h2=875m t=29s}

Answer 1

a) initial velocity =

angle =

horizontal velocity = V Cos

vertical velocity = V sin

initial height = h1

height of mountain = h2

maximum height of the projectile = h2 - h1

max height =

=> h2 - h1 = V^2 sin^2 / 2g

=> V^2 sin^2 = 2g (h2 - h1)

=> sin^2 = 2g (h2 - h1) / V^2

=> sin = sqrt [2g(h2-h1) / V^2]

=> = sin^-1 sqrt [2g(h2-h1) / V^2]

b) -h1 = V sin t - 1/2 gt^2

=> V sin t = 1/2 gt^2 - h1

=>

=>   

Range =  V cos t ( we can use the value of theta from above equation)

d) v0=155 m/s h1=95m, h2=875m t=29s

for a part : = sin^-1 sqrt [2g(h2-h1) / V^2] = sin^-1 [2*9.8* (875-95)] / 155^2

=>   = 39.51 degrees

for b part :

= sin^-1 ( 1/2*9.8 * 29^2 - 95) / 155*29

= 63.59 degrees

range = V cos t = 155* cos 63.59 * 29 = 1999.34 m