Ignoring air-resistance, equate energies:
(PE+KE)initial = (PE+KE)final
mgh+(1/2)mu² = (1/2)mv²
Solve for v.
I get approx. 71 m/s.
A cannon is fired from a cliff 200 m high downward at an angle of 30o with respect to the horizontal. If the muzzle velocity is 34 m/s, what is its speed (in m/s) when it hits the ground?
3 answers
thank you so much!
You're welcome!