Respuesta :
Answer:
4.5 m/s
Explanation:
The rock must barely clear the shelf below, this means that the horizontal distance covered must be
[tex]d_x = 5 m[/tex]
while the vertical distance covered must be
[tex]d_y = 6 m[/tex]
The rock is thrown horizontally with velocity [tex]v_x[/tex], so we can rewrite the horizontal distance as
[tex]d_x = v_x t[/tex]
where t is the time of flight. Re-arranging the equation,
[tex]t=\frac{d_x}{v_x}[/tex] (1)
The vertical distance covered instead is
[tex]d_y = \frac{1}{2}gt^2[/tex]
where we omit the term [tex]ut[/tex] since the initial vertical velocity is zero. From this equation,
[tex]t=\sqrt{\frac{2d_y}{g}}[/tex] (2)
Equating (1) and (2), we can solve the equation to find [tex]v_x[/tex]:
[tex]\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[/tex]