Projectile Motion Cannon Question

[NBAJam100]

Im new to the forum, so.. hello everyone!

I have a quick question on projectile motion that i can't seem to get...

A cannon 60m from the base of a 25m high cliff shoots a 15kg projectile at 43 degrees above the horizon...

(a)- what is the minimum muzzle velocity for the shell to clear to top of the cliff...

now i know that in the x direction- t= 60m/Vo(cos(43)t
In the y direction- 25m=Vo(cos(43))t+(.5)(9.8)t^2

I tried about 20 times to solve for either t or Vo and then plug that into the other equation and solve for one of the variables and then put that value into the next formula and solve for that but i can't seem to get the correct answer, which is apparently Vo= 32.6m/s. Where am i going wrong here?? Thanks!

 

[hage567]

now i know that in the x direction- t= 60m/Vo(cos(43)t
In the y direction- 25m=Vo(cos(43))t+(.5)(9.8)t^2

I don't know if you're just making typos here, but there shouldn't be a t at the end of your first equation.

In your second one, that should be sin43, not cos43, since it is the y component of the initial velocity. Also, watch your signs.

For future reference, there is a Homework Forum where you can post homework problems.

 

[NBAJam100]

Ah! yeah i don't know why i had those extra variables in there... ill be sure to post in the homework help section from now on, thanks! Got the answer by the way.