First of all let us find out the angle #theta_m # of the mountain peak from the location of enemy war ship
#theta_m= tan^-1 (1800/2500)approx 35.8^@#. It is clear that enemy war ship must fire at an angle #theta_m# so as to always clear the mountain peak and hit the ship (B) during the time of flight of the canon ball.
Let's assume that the canon ball is fired with its maximum speed only.
Velocity along #x#-axis #= u_x = u cos theta#
Acceleration along #x#-axis #a_x = 0#
Velocity along #y#-axis #= u_y = u sintheta#
Acceleration along #y#-axis #a_y = -g#
Different equation of motions of one dimension are:
#\vec{v}=\vecv_{0}-\vec{g}t# …... (1)
#\vec{y}-\vecy_{0}=\vecv_{0}t-\frac{1}{2}\vec{g}t^{2}# …... (2)
#v^2 = v_0^2 – 2g (y-y_0)# …... (3)
To obtain total Time of flight:
When body returns to the same horizontal level, the resultant displacement in vertical #y#-direction is zero. Use equation (2).
#0 = (u sin theta) t - 1/2g t^2#,
As #t# cannot equal to zero, total time of flight,
#t=2usintheta/g#.........(4)
Horizontal Range# = "Horizontal velocity" xx "Time of flight"#
#R= u cos theta xx 2 u sin theta/g#
#=(u^2 sin 2theta)/g#
Maximum range#=u^2/g =(250)^2/9.81=6371m#, For #sin 2theta=1#
Desired range is #"Distance of war ship from the mountain" +## "Distance of harbor from the mountain peak"=2500+300## =2800m#.
Angle #theta_d# required for this desired range,
Range #2800=(u^2 sin 2theta_d)/g=(250)^2/9.81 sin 2theta_d#
#sin 2theta_d=2800xx9.81/(250)^2#
#2theta_d=sin^-1(0.439488)=26.07#
#theta_dapprox13^@#
Now we also know that if angle of firing canon is #90-13=77^@#, the range will still be same, i.e., #2800m#.
Canon must be fired either at an angle of #13^@ or 77^@#
- Comparing with the angle of the mountain seen by the enemy war ship, when firing at and angle of #13^@# the mountain will provide natural obstruction.
- For firing angle of #77^@# the enemy war ship will not have any knowledge of ship's (B) location to take an indirect hit.
Therefore, the ship in harbor will always be safe.
