Question

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)

Answer 1

$v_y=gt,$

$h=\frac{gt^2}{2},\implies$

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

$v_y=vsin \beta,$

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

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