Question

A projectile is fired at an upward angle of 25.0^\circ from the top of a 310 m cliff with a speed of 220 m/s.
What will be its speed when it strikes the ground below? (Use conservation of energy.)

Answer 1

Assuming there is no air resistance, you can use the formula x(t)=x0+v0*t+1/2*a(t)^2 in order to find the time(t) it takes for the projectile to reach the ground below the cliff
0=310+220sin(25)*t-9.81/2*(t)^2

After finding the time, you can use the formula Vf=Vi+at

Vf=220-9.81*t

Then you're done

Answer 2

it says to use conservation of energy

we know conservation of energy when not dealing with heat is u+k=u+k where u is the potential energy and k is you kinetic energy. potential energy is mass x gravity x height. kinetic energy is (1/2)(mass)(velocity)^2.

from there you get the same answer as previous but in the proper method.