Question

You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 6 m at your feet, then a horizontal shelf of 5 m , then another drop of 4 m to the bottom of the canyon, which has a horizontal floor. You kick a 0.97 kg rock, giving it an initial horizontal velocity that barely clears the shelf below. 5 m ∆x 10 m v 6 m 4 m What initial horizontal velocity v will be required to barely clear the edge of the shelf below you? The acceleration of gravity is 9.8 m/s 2 . Consider air friction to be negligible. Answer in units of m/s.\

Answer 1

Answer:

4.5 m/s

Explanation:

The rock must barely clear the shelf below, this means that the horizontal distance covered must be

$d_x = 5 m$

while the vertical distance covered must be

$d_y = 6 m$

The rock is thrown horizontally with velocity $v_x$, so we can rewrite the horizontal distance as

$d_x = v_x t$

where t is the time of flight. Re-arranging the equation,

$t=\frac{d_x}{v_x}$ (1)

The vertical distance covered instead is

$d_y = \frac{1}{2}gt^2$

where we omit the term $ut$ since the initial vertical velocity is zero. From this equation,

$t=\sqrt{\frac{2d_y}{g}}$ (2)

Equating (1) and (2), we can solve the equation to find $v_x$:

$\frac{d_x}{v_x}=\sqrt{\frac{2d_y}{g}}\\\frac{d_x^2}{v_x^2}=\frac{2d_y}{g}\\v_x = d_x \sqrt{\frac{g}{2d_y}}=5\sqrt{\frac{9.8}{2(6)}}=4.5 m/s$