physics problem?

Question

A ship maneuvers to within 2500 m of an island's 1800 m high mountain peak and fires a proectile at an enemy ship 610 m on the other side of the peak. If the ship shoots the projectile with an initial velocity of 250 m/s at 75 degress, how close vertically) does the projectile come to the peak?

the ansewr is 210 m, but I don't know how to arrive this answer. So please explain this to me step by step if you can. Thank you.

Answer 1

Horizontal velocity = 250 x cos(75) = 64.7 m/s (constant)

Vertical velocity = 250 x sin(75) = 241 m/s. This also equals g * t so you can solve for the time it takes to get to its maximum height, t = 241 / 9.8 = 24.6 seconds.

The height is given by 1/2 g * t^2 = 4.9 x 24.6^2 = 2965 meters.

24.6 is exactly half the time of the whole flight, so in that time the projectile travels 64.7 x 24.6 = 1592 meters horizontally (this makes the actual range of the object 3183 meters, meaning it will overshoot the target by 73 meters).

The mountain is located at 2500 meters into the flight, which takes the projectile 2500 / 64.7 = 38.6 seconds. This is 14 seconds after it reaches its maximum altitude and has started falling back to earth, covering a vertical distance of 1/2 g t^2 = 4.9 x 14^2 = 960 meters below its highest point of 2965 meters, or 205 meters above the mountain.

Answer 2

If you dismiss the wind resistance (you don't have any data on it anyway)

You then know the path the projectile follows is a paraboloid trajectory. You know the sea level on both sides are at the same altitude. So you have 2 points on the path at a distance of 2500+610 meters apart. You also know the projectile leaves the nuzzle at a 75 degree angle. That should define the path the projectile takes.

x being the distance from the top:

using the two point where we hit the water:
altitude = y = (x+610)(x-2500).A
or y = Ax^2 -1890Ax -610.2500.A

determining A:
The angle in x=2500 is known to be 75 degrees (actually backwards) in my axis (make the drawing to figure it out).

dy/dx= 2Ax-1890A and for x=2500 dy/dx needs to be -tan(75degrees) or -3.7320

-3.7320=5000A-1890A
A= -3.7320/(5000-1890)

Filling that back in in the previous equation and solving it for the peak (x=0) give the altitude at the peak as (+610)(-2500).-3.7320/(5000-1890) or 1830 meters.

I'm sorry to conclude the projectile would only surpass the peak with approximately 30 meters. So I'm not sure where you got the 210 meter answer.

The highest point on the path is reached where dy/dx=0 or where x=1890/2=945 and the altitude there is (945+610)(945-2500).-3.7320/(5000-1890) or 2901.63 meters

Answer 3

taking the x axis movement from s=ut+.5at^2
in x axis a=0 so eqn reduces to s=ut.
u=250 cos 75 so we find t from it to b t=2500/(250cos75)=38.637
now using y axis eqn.
s=ut +0.5at^2. here u = 250 sin75 t=38.637 and a=-9.8m/s
so s=250sin75*38.637-0.5*9.8*t^2 so s=2015.312
so distance between cliff top and shot is j=2015.312-1800=215.312( near to your answer. maybe not rounding much)
hope this helps