A stone is thrown at 15 m/s an angle \theta below the horizontal from a cliff of height H. It lands 78 m from the base 6 s later. Find \theta and H.

Answer 1

Vo = 15m/s @ A Degrees.
Xo = 15*Cos A.
Yo = 15*Sin A.

Dx = Xo*Tf = Xo * 6 = 78 m.
Xo = 13 m/s.

Xo = 15*Cos A = 13 m/s.
15*Cos A = 13
Cos A = 13/15 = 0.86667
A = 29.93o

Yo = 15*Sin29.93 = 7.48 m/s.

h = Yo*T + 0.5g*t^2 = 7.48*6 + 4.9*6^2
= 221.3 m.