Question

A man stands on the roof of a building of height y0 and throws a rock with a velocity of magnitude v0 at an angle of ? above the horizontal. You can ignore air resistance. 1)Calculate the magnitude of the velocity of the rock just before it strikes the ground. Take free fall acceleration to be g. 2) Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground. Take free fall acceleration to be g

Answer 1

See citation. Assuming meters and a = -9.8 m/s,
Height = (V^2*sin^2A)/2g = (Vo^2*sin^2(theta))/2g m
Range = (V^2 sin2A)/g = (Vo^2 * sin(2theta)) / 9.8 m
But this is at Yom. The rock's velocity is now Vo at an angle of theta BELOW the horizontal.
Vertical component of velocity is V0*sintheta
We can calculate how long it takes to reach the ground from here
s = Vo*t + 0.5*a*t^2
t=sqrt(s-Vo*t /0.5*a)
only the positive root makes sense, so
The horizontal distance traversed in this time is
s = Voh*t
This gets added to R calculated above, so
Total horizontal distance = t+ s
The vertical velocity at this time is
Vv = Vo +a*t
The horizontal velocity is unchanged at Vo*costheta
So the magnitude of velocity at impact = sqrt(horizontal velocity^2 + Vv^2)