physics question?

Question

An enemy ship is on the east side of a mountain island. The enemy ship has maneuvered to within 2 500 m of the 1 800-m-high mountain peak and only can shoot projectiles with an inital speed of 250 m/s. If the western shoreline is horizontally 300 m from the peak, what are the distances from the bombardment of the enemy ship?

Answer 1

OK, this was complicated.

First, I calculated the angle of the enemy gun just to hit the target, forgetting the mountain using flight time

300+x=cos(a)*v*t

where a is the angle of the gun
and v is the initial speed
x is the distance from the island
t=(300+x)/(cos(a)*v)

Now, the flight time will be twice the time to apogee in the vertical:

1/2 *t *g=sin(a)*v

t=2*sin(a)*v/g

the time is the same, so
2*sin(a)*v/g=
(300+x)/(cos(a)*v)

Solve for the angle
sin(a)*cos(a)=
g*(300+x)/(2*v^2)

sin(a) = 1-cos(a)
so
cos(a)^2-cos(a)+
g*(300+x)/(2*v^2)
=0

find the two roots. Both angles will work.

But is the angle sufficient to clear the mountain?

This is where it gets interesting

The projectile must have an altitude greater than 1800 m when it flies over the mountan.

As long as the ship is greater than 300 m from the mountain, it must clear the mountain on the descent. I will assume this is always the case by assuming the mountain is centered on the island.

The angle that will work for this is
ACos(1-
sqrt(1-4*g*(300+x)/
(2*v^2))/2)
call this Acos(K)
where K is the expression

Relating the position of the projectile to the angle that it was fired after apogee
y=sin(a)*v*t-.5*g*t^2
=(1-cos(a))*v*t-.5*g*t^2
=(1-K)*v*t-.5*g*t^2
t is greater than sin(a)*v/g

and x=v*cos(a)*t
or
x=v*K*t
t=x/(v*K)

plug into the equation for y
y=
=(1-K)*v*(x/(v*K))-
.5*g*(x/(v*K))^2

Note that y is now expressed in terms of X
and all other variables are known since the y we care about is 1800

So, I looked at where the height is 1800 or greater.

turns out that at 978 m from the mountain, is the first time the projectile will clear the peak and hit the target.

I used excell to set up the equations and then iterate on the answer. The flight time is 45.2 sec
the angle is 83.507 degrees.

There is likely some elegant way to arrive at the answer. Starting from my work, write out y set to 1800 and solve for x. The algebra alone will take alot of care.

The ship can continue to fire using the angle and range from the shore equation I derived above as long as it stays within 978m of the mountain.

j