A cannon-ball is fired horizontally off of the edge of a cliff. When the cannon-ball hits the water below it hits the water with a velocity of 19 m s−1 at an angle of 65° below the horizontal.
How tall was the cliff? (in m to 2.s.f)
vy=gt,v_y=gt,vy=gt,
h=gt22, ⟹ h=\frac{gt^2}{2},\impliesh=2gt2,⟹
h=vy22g,h=\frac{v_y^2}{2g},h=2gvy2,
vy=vsinβ,v_y=vsin \beta,vy=vsinβ,
h=v2sin2β2g,h=\frac{v^ 2 sin^2\beta}{2g},h=2gv2sin2β,
h=192⋅sin265°2⋅9.81≈15.11 m.h=\frac{19^2\cdot sin^2 65°}{2\cdot 9.81}\approx15.11~m.h=2⋅9.81192⋅sin265°≈15.11 m.
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