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?
3 answers
Also find the final velocity in y (v_fy) of the shot just before it hits the ground.
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
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
Range = Vo^2*sin(2A)/g.
Range = 65^2*sin(2*37)/9.8 = 414.4 m.
Range = 65^2*sin(2*37)/9.8 = 414.4 m.
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