Projectile question — Cannonball fired over a hill

In summary: The initial velocity should be greater than the one previously calculated for vertical shooting. The initial velocity should be greater than 71.42 m/s.
  • #1
dyn
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Homework Statement
A 130m high hill is located halfway between the cannon and its target. If the cannonball is fired at an elevation angle of 30 degrees and just clears the hill to strike the target , calculate the initial velocity and the distance from the cannon to the target
Relevant Equations
Time of flight , t= 2v sin(theta) /g where v is the initial speed and theta is the elevation angle
Time of flight is t = 2v sinθ /g = v/g for θ = π/6 so time to top of flight is t = v/(2g)

I then constructed a right-angle triangle with θ=π/6 and opposite side of 130m and used tangent to get the adjacent side , ie. distance from cannon to hill to be 225.17m
The horizontal velocity is v cosθ which is 0.866v. Using speed = distance /time I equated 0.866v to 225.17/ {v/(2g)} to get an equation for v squared which gives a final answer of v = 71.42 m/s and distance from cannon to target is 2 x 225.17 = 450.34 m

These are not the right answers but I have no idea why ?
 
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  • #2
The mistake is here:
dyn said:
I then constructed a right-angle triangle with θ=π/6 and opposite side of 130m and used tangent to get the adjacent side , ie. distance from cannon to hill to be 225.17m
Make a sketch in which you draw the trajectory of the cannonball from the cannon to the top of the hill. (What is the shape of the trajectory?) Then draw a straight line from the cannon to the top of the hill. Does the straight line make a 30o angle to the horizontal?
 
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  • #3
Thanks for your reply. The trajectory is a parabola so I can't use the right angle triangle
 
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  • #4
Try this approach: you are told the highest vertical distance of the projectile. How long would it take to reach this height?
 
  • #5
dyn said:
Thanks for your reply. The trajectory is a parabola so I can't use the right angle triangle
What the initial velocity should be if the cannon is pointed perfectly vertical and you want the cannonball to reach the height of 130 meters only?
Then, if you tilt the cannon to an elevation angle of 30 degrees, trying to cover the horizontal distance to the target, still clearing the hill, should the initial velocity be greater than the one previously calculated for vertical shooting?
If so, how much it and its horizontal component should be?
 
  • #6
Thanks everyone. I got it right now.
 
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1. What is a projectile?

A projectile is any object that is launched or thrown into the air and is affected by the force of gravity.

2. How does a cannonball fired over a hill behave?

A cannonball fired over a hill follows a curved path known as a parabola. This is because the force of gravity pulls the cannonball towards the ground while the initial velocity from the cannon gives it a forward motion.

3. What factors affect the trajectory of a cannonball fired over a hill?

The trajectory of a cannonball fired over a hill is affected by the initial velocity, the angle of elevation, the force of gravity, and air resistance. The shape and weight of the cannonball can also play a role.

4. How can we calculate the range of a cannonball fired over a hill?

The range of a cannonball fired over a hill can be calculated using the formula: R = (v^2 * sin(2θ)) / g, where R is the range, v is the initial velocity, θ is the angle of elevation, and g is the acceleration due to gravity.

5. What happens to the cannonball after it reaches its highest point?

After reaching its highest point, the cannonball will start to fall back towards the ground due to the force of gravity. It will follow the same parabolic path until it reaches the ground, assuming there are no other external forces acting on it.

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