- #1
EvanQ
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Homework Statement
A cannon, located 60.0m from the base of a vertical 25.0m tall cliff, shoots a 15 kg shell at 43.0degrees above the horizontal toward the cliff.
What must the minimum muzzle velocity be for the shell to clear the top of the cliff?
The ground at the top of the cliff is level, with a constant elevation of 25.0 m above the cannon. Under the conditions of part A, how far does the shell land past the edge of the cliff?
Homework Equations
2-dimensional motion equations such as s=ut+(1/2)at^2 etc. also tried using the range formula 2U/g sin(theta) cos(theta).
The Attempt at a Solution
i've been stuck on this problem for hours and feel like I've exhausted all of the options that i understand / am aware of. i initially tried to get a reading of time by calculating how long it would take for the ball to drop the 25m, and came out with approx 2.259, then used s=ut+(1/2)at^2 to calculate that the vertical velocity needed to make the 25m was 22.136m/s. But then after rearranging that to find the actual muzzle velocity, i think i found that that was insufficient to make the 60metres. so then i tried tackling it using that range formula i listed, and came out with a muzzle velocity of 25.23m/s, which out of frustration i entered to find i was incorrect. i have 2 more submission attempts and i am really confused as to how to attack this problem. any help would be greatly appreciated.
-evan.