Posted by jm on .
a man stands on the roof of a building that is 30m tall and throws a rock with a velocity of magnitude of 40m/s at an angle of 33 degress a bove the horizontal. Calculate the following:
a.) The maximum height above the roof reached by the rock.
b.) The magnitude of the velocity of the rock just before it strikes the ground.
c.) The horizontal distance from the base of the building to the point where the rock strikes the ground.
Vertical initial velocity = Vyo
= 40 sin33 = 21.79 m/s
Horizontal velocity = Vx
= 40 cos 33 = 33.55 m/s
a) gH = (1/2) Vyo^2
H = 24.22 m
b) gH + Voy^2/2 + Vx^2/2 = (1/2)Vfinal^2
(1/2)Vfinal^2 = 237.4 + 474.8 + 562.8
= Vfinal = 46.37 m/s
c) Vx*(time of flight)
To get the time of flight, t, solve
(-g/2)t^2 + Vyo*t = -24.22 m