A projectile is fired at an upward angle of 55.0 degrees from the top of a 315m- cliff with a speed of 160 m/s .What will be its speed when it strikes the ground below? (Use conservation of energy.)
So far I have gotten: K1+Ug1+Us1-mew fnd=k2+Ug2+Us2 and I have crossed out Us1 and Us2 since there is no spring involved and I assume that there is a kinetic energy 1 and 2 and gravity is involved as well. I am confused on how to apply the angle to this problem. and how to set the rest of the problem up! Thank you for any help :)

Question

A projectile is fired at an upward angle of 55.0 degrees from the top of a 315m- cliff with a speed of 160 m/s .What will be its speed when it strikes the ground below? (Use conservation of energy.) 
So far I have gotten: K1+Ug1+Us1-mew fnd=k2+Ug2+Us2 and I have crossed out Us1 and Us2 since there is no spring involved and I assume that there is a kinetic energy 1 and 2 and gravity is involved as well. I am confused on how to apply the angle to this problem. and how to set the rest of the problem up! Thank you for any help :)

anonymous · Answer

is the answer 23352 m/s?

anonymous · Answer

@Tushara : ans seems too much in magnitude..
@MartialArtGirl647:
since speed is involved not velocity so u don't need the angle condition at all..
the initial energy of projectile is [ kinetic+gravitational potential] energy..all of this energy finally converts to the final Kinetic energy at ground..|dw:1339937601988:dw|  where
u=160m/s, h=height of cliff ,g=9.8m/s^2..by calculation
v=final speed=178.25m/s