A projectile is fired into the air from the top of a cliff of height h = 186 m above a valley. Its initial velocity is v0 = 64.8 m/s at an angle θ = 59° above the horizontal. Where does the projectile land? (Ignore any effects due to air resistance.)

Answer 1

ho = 186 m.
Vo = 64.8m/s[59o]
Xo = 64.8*Cos59 = 33.37 m/s
Yo = 64.8*sin59 = 55.54 m/s

Tr = -Yo/g = -55.54/-9.8 = 5.67 s. = Rise time.

h = ho + (-Yo^2)/2g
h = 186 + -(55.54^2)/-19.6 = 343.4 m. Above gnd.

h = 0.5g*t^2 = 343.4 m.
4.9t^2 = 343.4
t^2 = 70.1
Tf = 8.37 s. = Fall time.

d = Xo*(Tr+Tf) = 33.37m/s * (5.67+8.37)s
= 469 m. = Hor. distance.