Solving Motion in 2 Dimensions: Hammer Dropped from 10m Roof

[Beretta]

A worker on the roof of a house drops her hammer which slides down the roof at constant speed of 4m/s. Th roof makes an angle of 30 with the horizontal, and the lowest point is 10m from the ground. what is the the horizontal distance traveled by the hammer between the time is leaves the roof of the house and the time it hits the ground?

What happen here is that I assumed that theta is 60 since it makes an angle if 30 with the horizontal and since the object is sliding down. Thus it leaves the roof on a 60 degree angle.

vx = 4m/s cos 60 = 2m/s
vy = 4m/s sin 60 = 3.46m/s

x(t) = x0 + vxt

y(t) = y0 + v0t - (1/2)g(t^2)
10m - 3.46m/s(t) + (0.5)(9.81m/s^2)t^2 = 0

I am ending up with a negative root Delta.
May you help me pls?

 

[LightHero]

The horizontal distance traveled by the hammer is found using the x equation: x(t) = x0 + vxt. Since the hammer is initially at the roof, x0 = 0 and vx = 2m/s, so the equation becomes x(t) = 2m/s t. To find t, we solve the y equation for t: 10m - 3.46m/s(t) + (0.5)(9.81m/s^2)t^2 = 0. This gives us t = 1.345 seconds. So, the horizontal distance traveled by the hammer is x(t) = 2m/s * 1.345s = 2.69 m.