consider the motion along the X-direction
X = horizontal displacement = 80 m
[tex]V_{ox}[/tex] = initial velocity along the x-direction = v Cos60
t = time of travel
using the equation
X = [tex]V_{ox}[/tex] t
80 = (v Cos60) (t)
t = 160/v eq-1
consider the motion in vertical direction :
Y = vertical displacement = 20 m
[tex]V_{oy}[/tex] = initial velocity in Y-direction = v Sin60
a = acceleration = - 9.8 m/s²
t = time of travel = 160/v
using the equation
Y = [tex]V_{oy}[/tex] t + (0.5) a t²
20 = (v Sin60) (160/v) + (0.5) (- 9.8) (160/v)²
v = 32.5 m/s
