Projectile question — Cannonball fired over a hill

[dyn]

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 ?

 

[TSny]

The mistake is here:

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 30^o angle to the horizontal?

 

[dyn]

Thanks for your reply. The trajectory is a parabola so I can't use the right angle triangle

 

[scottdave]

Try this approach: you are told the highest vertical distance of the projectile. How long would it take to reach this height?

 

[Lnewqban]

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?

 

[dyn]

Thanks everyone. I got it right now.