James Bond Mission: Find Height of Cannon from Sea

  • Thread starter Thread starter maxtheminawes
  • Start date Start date
  • Tags Tags
    Bond
AI Thread Summary
James Bond's cannonball is shot at a 40-degree angle with an initial velocity of 35 m/s, landing 150 m from the cliff base. The time of flight is calculated using horizontal motion, yielding approximately 5.59 seconds. The vertical height is then determined using the projectile motion equations, resulting in a height of 27.35 m. However, a sign error is noted, indicating the height should be -27.35 m, reflecting the downward direction of the projectile. The discussion emphasizes the importance of correctly resolving velocity components and applying the appropriate signs in calculations.
maxtheminawes
Messages
20
Reaction score
0

Homework Statement


James Bond job is to aim at clifftop cannons at attacking vessels coming from the sea. In one practice exercise he shoots her cannon at an angle of 40 degrees, giving the cannon ball a velocity of 35m/s. If the cannon ball lands 150m from the base of the cliff, how high is James abouve the sea.


Homework Equations


v=vo+at
y=yo+vot+1/2at^2
y=height

The Attempt at a Solution


v=vo+at
35=(-9.8)t
t=3.57secs

y=yo+vot+1/2at^2
y=35sin40(3.57)+1/2(-9.8)3.58^2
y=17.87m

 
Physics news on Phys.org
Are you familiar with projectile motion? Try splitting the attack into two parts, where the cannon ball is rising (-ve g) and where it falls (+ve g). Resolving your velocity into components will give you a good start.
 
Alternately, since the horizontal velocity of the cannon ball is constant in time (no air drag), you can compute the time of flight of the shot. Knowing that and the initial vertical velocity component, you can compute the vertical distance by using the projectile motion equation for y. Just be careful with the signs in the equation.
 
ok, i retried the problem. is this right?
x=vt
150=35cos40t
t=5.59 sec

y=Voyt+1/2at^2
y=35sin40(5.59)+1/2(-9.8)(5.59^2)
y=27.35m
 
Last edited:
maxtheminawes said:
ok, i retried the problem. is this right?
x=vt
150=35cos40t
t=5.59 sec

y=Voyt+1/2at^2
y=35sin40(5.59)+1/2(-9.8)(5.59^2)
y=27.35m

That's it essentially but you have a sign error. y = -27.35 based on your equation.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top