Answer:
The rate of ladder moving down against the wall is 1.125 ft/s
Explanation:
Note: Refer to the attached figure
Applying the concept of the Pythagoras we have
[tex]x^2+y^2 = H^2[/tex]
or
9²+y² = 15²
⇒y² = 225 - 81
⇒y = √144 = 12 m
also,
[tex]x^2+y^2 = H^2[/tex]
differentiating with respect to time, we get
[tex]2x\frac{dx}{dt}+2y\frac{dy}{dt} = 0[/tex]
substituting the values in the above equation we get
[tex]2\times 9\timesfrac1.5+2\times 12\frac{dy}{dt} = 0[/tex]
or
[tex]27+24\frac{dy}{dt} = 0[/tex]
or
[tex]\frac{dy}{dt} =-\frac{27}{24}[/tex]
or
[tex]\frac{dy}{dt} =-1.125\ ft/s[/tex]
here, the negative sign depicts that the value of y is decreasing or the ladder is moving down.
hence, the rate of ladder moving down against the wall is 1.125 ft/s
Answer:1.125 ft/sec
Explanation:
Given
length of ladder =15 ft
Base of ladder is 9 m away
Angle which ladder makes with horizontal is
[tex]cos\theta =\frac{9}{15}[/tex]
Let at any instant base of ladder is x m away and top of ladder is ym away from ground
therefore
[tex]x^2+y^2=L^2[/tex]
differentiating w.r.t we get
[tex]x\frac{\mathrm{d} x}{\mathrm{d} t}+y\frac{\mathrm{d} y}{\mathrm{d} t}=0[/tex]
[tex]x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}[/tex]
[tex]x\times \dot{x}=-y\times \dot{y}[/tex]
[tex]\frac{x}{y}\times \dot{x}=-\dot{y}[/tex]
[tex]\dot{y}=\frac{3}{4}\times 1.5=\frac{9}{8}ft/sec[/tex]
