How far does the shell land past the edge of the cliff? (projectile motion)?

Question

Here is the entire question:

A canon, located 60 meters from the base of a vertical 25 meter tall cliff, shoots a 15 kilogram shell at 43 degrees above the horizontal toward the cliff.

I found the minimum muzzle velocity for the shell to clear the top of the cliff to be 32.6 m/s.

Under these conditions, how far does the shell land past the edge of the cliff?


Thanks a lot for your time!

Oh yeah, I forgot to mention that the ground at the top of the cliff is level, with a constant elevation of 25.0 m above the cannon... that might make a difference.

Answer 1

Your solution lands the shell exactly on the lip of the cliff.
BTW: I used g=9.81 and got 32.7 (32.66) m/s

Since
y(t)= 32.7*sin(43)*t-.5*9.81*t^2

solve for t when y(t)= 25.

You will find t=2.03
and t=2.51

When t= 2.03, x(t)=48.5, which means the cannon ball has not traveled the distance to the cliff.

When t=2.51, x(t)=60, which means the cannon ball was descending when it reached the lip of the cliff. In order to land beyond the lip, v0 must be increased. Then the t of landing will always be

(v0*sin(43)+
sqrt(v0^2*sin^2(43)-490.5))/
9.81

then
x(t)=v0*cos(43)*(v0*sin(43)+
sqrt(v0^2*sin^2(43)-490.5))/
9.81

so choose a v0 greater than 32.7 and plug and chug.

And, the distance beyond, lets call s is
s=60-x(t)

j