You know its initial PE and KE. At the bottom of the cliff, it only has finalKE.
1/2 m vf^2=mg*265 + 1/2 m 185^2
solve for vf
A projectile is fired at an upward angle of 45 degrees from the top of a 265-m cliff with a speed of 185 m/s. What will be its speed when strikes the ground below? (Use conservation of energy.)
I have a diagram of it but I don't know what equations to use. Help?
5 answers
Does that mean the answer would be the square root of 39419?
I get that.
Okay thanks.
If you use the kinematic equation Vf^2=VoSin 45 + 2gd the answer would differ from one of the conservation of energy equation method.
Vo=185m/s and vf=0 at max height
Why is this?
Vo=185m/s and vf=0 at max height
Why is this?
2 answers