- #1

dyn

- 749

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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 ?

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 ?