Answer to Question #103653 in Mechanics | Relativity for L3M0N555

Question

santa claus is delivering christmas presents to the children of venice. a venice house has a roof that slopes 60, with its lowest edge above the water of a channel. as santa exits the chimney, he is unaware that some misguided citizen has coated the roof in frictionless snow to make him feel at home. her starts to slide down the roof, travelling 3.77m to the lower edge. at that instant, frightened for his life, santa jumps with a speed of 6 m/s at an angle of 90 relative to the rooftop. if the roof edge is 10.06 above the channel how far from the house does santa splash land

Answer 1

Calculate his final speed at the lower edge of the roof that he achieved while was moving down:

$$v_f=\sqrt{2gh}=\sqrt{2gL\text{ sin}60^\circ}=8\text{ m/s}.$$

As we can guess, this speed is directed 60 degrees below the horizontal. This speed adds (as a vector) to the speed $u=6$ m/s that is vertical to the roof surface. The magnitude of the resultant speed is

$$v=\sqrt{v_f^2+u^2}=10\text{ m/s}.$$

It is directed

$$\theta=\text{atan}\frac{u}{v_f}=\text{atan}\frac{6}{8}=36.9^\circ$$

above $v$ vector.

Therefore, the resultant speed $v$ is directed

$$\alpha=180^\circ-\theta-120^\circ=23.1^\circ$$

below the horizontal. See the figure:

Now we know that the vertical and horizontal components of the speed is

$$v_v=v\text{ sin}\alpha=3.92\text{ m/s},\\ v_h=v\text{ cos}\alpha=9.2\text{ m/s}.$$

The time for motion vertically can be found from the following equation:

$$H=y_0+vt+\frac{gt^2}{2}.$$

Solve for time:

$$t=\frac{\sqrt{v^2+2gH}-v}{g}=1.09\text{ s}.$$

Meanwhile the object traveled horizontally at a constant speed:

$$R=v_ht=10.03\text{ m}.$$