Answer to Question #326211 in Mechanics | Relativity for Muhammad

Question

Calculations:

1. Suppose a large rock is ejected from a volcano with a speed of 25m/s and at an angle of 35*

the horizontal. If the rock strikes the side of the volcano at an altitude 20m lower than the star

point.

(a) Calculate the time it takes the rock to reach the starting point.

(b) What are the magnitude and direction of the rock's velocity at impact?

Answer 1

$Let\space v_0=25\space ms^{-1},\space\alpha_0=35\degree,\\ h=20\space m,\space g=9.8\space ms^{-2}$

a) Let's write an equation for vertical direction:

$y=v_0\sin(\alpha_0)t-\frac{gt^2}{2}\\ y=-h\Rarr\frac{gt^2}{2}-v_0\sin(\alpha_0)t-h=0\\ t=\frac{v_0\sin\alpha_0\pm\sqrt{(v_0\sin\alpha_0)^2 + 2gh}}{g}$

Since time should be $\geq0$ , we select '+' sign:

$t=\frac{v_0\sin\alpha_0+\sqrt{(v_0\sin\alpha_0)^2 + 2gh}}{g}$

$t=\frac{25\cdot\sin{35\degree}+\sqrt{(25\cdot\sin{35\degree})^2 + 2\cdot 9.8\cdot 20}}{9.8}\space s\approx 4.0\space s$

b) We can find velocity magnitude at impact from the law of conservation of energy:

$\frac{mv_0^2}{2}=\frac{mv^2}{2}-mgh\Rarr v=\sqrt{v_0^2+2gh}\\ v=\sqrt{25^2+2\cdot 9.8\cdot 20}\space ms^{-1}\approx 32\space ms^{-1}$

The angle between velocity and horizontal we can find using fact that horizontal component of speed remains constant:

$v_x=v_0\cos\alpha_0=const\\ \cos\alpha=\frac{v_x}{v}=\frac{v_0\cos\alpha_0}{\sqrt{v_0^2+2gh}}\Rarr \alpha=\pm\arccos{\frac{v_0\cos\alpha_0}{\sqrt{v_0^2+2gh}}}$

Since velocity is directed downward, we select '-' sign:

$\alpha=-\arccos\frac{25\cdot \cos{35\degree}}{32}\approx -50\degree$

Answers:

a) $t\approx 4.0\space s$

b) $v\approx 32\space ms^{-1},\space \alpha\approx -50\degree$