Shooting a cannonball over a cliff

[clarineterr]

Homework Statement

A cannon 60m horizontally from the base of a 25m cliff is fired with a speed of 32.6m/s at an angle of 43 degrees above the horizontal towards the cliff. How far does the shell land past the edge of the cliff assuming its elevation is constant

Homework Equations

y = v0yt -.5gt^2

x= v0xt

The Attempt at a Solution

I plug in the to the first equation

25m = 32.6sin(43)t -4.9t^2

and the quadratic formula gives me two answers: 2.46s and 2.06s

so I plug this into the second equation giving me distances of 49.0m and 58.6

which seems to mean that the cannonball does not actually clear the cliff...but my online how system is asking for an answer.

 

[LowlyPion]

Homework Statement

A cannon 60m horizontally from the base of a 25m cliff is fired with a speed of 32.6m/s at an angle of 43 degrees above the horizontal towards the cliff. How far does the shell land past the edge of the cliff assuming its elevation is constant

Homework Equations

y = v0yt -.5gt^2

x= v0xt

The Attempt at a Solution

I plug in the to the first equation

25m = 32.6sin(43)t -4.9t^2

and the quadratic formula gives me two answers: 2.46s and 2.06s

so I plug this into the second equation giving me distances of 49.0m and 58.6

which seems to mean that the cannonball does not actually clear the cliff...but my online how system is asking for an answer.

If it gives 2 answers, isn't one when it passes the height going up and the other when it lands?. So don't you first need to figure what the horizontal velocity is? Then take the longer time and figure the horizontal distance from the gun (2.46*32.6*cos(43)) then subtract the distance from the base of the cliff?

Edit: I see. Your solutions are not reaching the cliff. Check your quadratic again or maybe enter -1.4 m then

 

[LowlyPion]

I get 2.44 sec and 2.09 sec. Which falls even a little shorter.

Are you sure you have the right statement of the problem?

As stated, I would have to wonder.

 

[tkahn6]

Homework Statement

A cannon 60m horizontally from the base of a 25m cliff is fired with a speed of 32.6m/s at an angle of 43 degrees above the horizontal towards the cliff. How far does the shell land past the edge of the cliff assuming its elevation is constant

The way I read this, the cannon is sitting at the bottom of a valley shooting a cannon ball towards a cliff that is 60m away and 25m high.

Therefore, shouldn't you be looking to see where the canon is vertically at t where x_horizontal = 60m?

I get about 24.887m at t = 2.51655 for x_vertical. Assuming this is just a rounding error, the shell would land 0m past the cliff's edge.

 

[negitron]

Therefore, shouldn't you be looking to see where the canon is vertically at t where x_horizontal = 60m?

That's going to happen at two points; one on the way up and one on the way back down. You found the first.

 

[ideasrule]

That's going to happen at two points; one on the way up and one on the way back down. You found the first.

In other words, it's true that the cannon ball just barely avoids hitting the cliff, but it's still moving after scraping by and will eventually hit the ground.