A small disc of mass \(m\) slides down with initial velocity zero from the top \((A)\) of a smooth hill of height \(H\), having a horizontal portion \((BC)\) as shown in the figure. If the height of the horizontal portion of the hill is \(h\), then the maximum horizontal distance covered by the disc from the point \(D\) is
Show Hint
For projectile motion after sliding from a smooth track, first use conservation of energy to find horizontal speed, then use vertical motion to find time of flight.
Step 1: Find the velocity of the disc at point \(C\).
The hill is smooth, so mechanical energy is conserved.
At point \(A\), the disc starts from rest. Therefore,
\[
K_A=0
\]
The height of point \(A\) from the ground is
\[
H
\]
The height of point \(C\) from the ground is
\[
h
\]
Loss in potential energy from \(A\) to \(C\) becomes kinetic energy at \(C\).
Hence,
\[
mgH-mgh=\frac{1}{2}mv^2
\]
\[
mg(H-h)=\frac{1}{2}mv^2
\]
Cancelling \(m\),
\[
g(H-h)=\frac{v^2}{2}
\]
Therefore,
\[
v^2=2g(H-h)
\]
Step 2: Time taken by the disc to fall from height \(h\).
At point \(C\), the disc leaves the horizontal portion with horizontal velocity \(v\).
The vertical height from point \(C\) to the ground is
\[
h
\]
For vertical motion,
\[
h=\frac{1}{2}gt^2
\]
Thus,
\[
t^2=\frac{2h}{g}
\]
\[
t=\sqrt{\frac{2h}{g}}
\]
Step 3: Horizontal distance from point \(D\).
The horizontal distance covered is
\[
R=vt
\]
Substituting the values,
\[
R=\sqrt{2g(H-h)}\times \sqrt{\frac{2h}{g}}
\]
\[
R=\sqrt{4h(H-h)}
\]
\[
R=2\sqrt{h(H-h)}
\]
Step 4: Find the maximum value of \(R\).
We need to maximize
\[
R=2\sqrt{h(H-h)}
\]
So, maximize
\[
h(H-h)
\]
Let
\[
f(h)=h(H-h)
\]
\[
f(h)=Hh-h^2
\]
This quadratic is maximum at
\[
h=\frac{H}{2}
\]
Therefore,
\[
R_{\max}=2\sqrt{\frac{H}{2}\left(H-\frac{H}{2}\right)}
\]
\[
R_{\max}=2\sqrt{\frac{H}{2}\cdot \frac{H}{2}}
\]
\[
R_{\max}=2\sqrt{\frac{H^2}{4}}
\]
\[
R_{\max}=H
\]
Step 5: Final conclusion.
Hence, the maximum horizontal distance covered by the disc from point \(D\) is
\[
\boxed{H}
\]
