Question

A projectile is shot from the edge of a cliff 145m above ground level with an initial speed of 65.0 m/s at an angle of 37.0° with the horizontal. How far did the shot go horizontally?

Answer 1

Also find the final velocity in y (v_fy) of the shot just before it hits the ground.

Answer 2

horizontal components
Vox=65cos(37)=51.11 = Vfx
verticall components
Voy=65sin(37)=39.11 ==> Vfy=-39.11
find t from this formula Vfy=Voy-gt and solve for t. t=7.98s
plug t in this equation X=Voxt+0.5(a)t^2
X=414.32

Answer 3

Range = Vo^2*sin(2A)/g.
Range = 65^2*sin(2*37)/9.8 = 414.4 m.