Answer: 50.4 m/s
Explanation:
height of the building = 130 m
horizontal distance = 64 m
acceleration due to gravity = 9.8 m/s
applying the equation of motion s = ut + (0.5at^2) and assuming that air resistance is neglected, we first find the time it takes to hit the ground
where s = distance (height of the building)
u = initial velocity (which is 0 in this case )
t = time
a= acceleration due to gravity
130 = ( 0 x t) + ( 0.5 x 9.8 x t^2)
130 = 4.9 x t^2
t = [tex]\sqrt{(130 / 4.9)} \\[/tex]
t = 5.15 secs
now we use the equation Vy = Uy + at to get its speed before hitting the ground
where Uy = initial velocity(0 in this case)
Vy = 0 + (9.8 x 5.15)
Vy = 50.4 m/s
