Answer to Question #175705 in Mechanics | Relativity for Hamaika Manuel

Question #175705

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)


1
Expert's answer
2021-03-26T11:38:00-0400

vy=gt,v_y=gt,

h=gt22,    h=\frac{gt^2}{2},\implies

h=vy22g,h=\frac{v_y^2}{2g},

vy=vsinβ,v_y=vsin \beta,

h=v2sin2β2g,h=\frac{v^ 2 sin^2\beta}{2g},

h=192sin265°29.8115.11 m.h=\frac{19^2\cdot sin^2 65°}{2\cdot 9.81}\approx15.11~m.


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