Watching your ship come in...You are stanidng on top of a building h = 100 f t h

Question

Watching your ship come in...You are stanidng on top of a building h = 100 f t high, watching a boat sail directly toward you. The angle between your line of sight (the line connecting your eye and the boat) changes from 1 = 25 to 2 = 40 during two minutes’ observation (t = 2.0 min). Find the boat's speed (v) in miles per hour.

...and then trying to sink it. The cannon atop your building has muzzle speed v = 1400 mi/hr.

(a) How far from your building will the cannonball land if you fire the cannon pointed = 30 above the horizon. Report your answer in miles.

(b) Find the angle between your line of sight and the horizon as you watch the cannonball splash into the ocean.

(c) If the boat stops and remains stationary 19.0 miles from you, What angle above the horizon should you point your cannon in order to hit it?

Explanation / Answer

displacement of the boat = dx = y*( (1/tan25 - 1/tan40)

dx = 100*( (1/tan25 - 1/tan40) = 95.3 ft = 0.018 mile

dt = 2min = 1/30 hr

speed = v = dx/dt = 0.018*30 = 0.54 miles/hr

(a)

for a projectile

y = x*tantheta - 0.5*g*x^2/(v^2*(costheta)^2)

y = 0.0189 mile

g = 78919.11 mile/hr^2

-0.0189 = (x*tan30) - 0.5*78919.11*x^2/(1400^2*(cos30)^2)

x = 21.54 miles

++++

(b)

vy^2 - (vo*sin30)^2 = 2*g*y

vy^2 - (1400*sin30)^2 = 2*78919.11*0.0189

vy = 702.13 miles/hr

vx = vox = 1400*cos30 = 1212.43 miles/hr

angle = tan^-1(vy/vx) = 30.7

++++

x = 19

-0.0189 = 19*tantheta - 0.5*78919.11*19^2/(1400^2*(costheta)^2)

theta = 24.88 degrees

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