How long does it take for the object to slide distance $l$ on the inclined plane?

Question

Consider the following problem:

Suppose than an object is allowed to slide down an inclined plane from a point which is $h$ feet above the horizontal ground, and that the object slides $l$ feet along the plane to the ground . How long does it take to slide this distance?

After trying to solve this myself, I found that the answer provided by the textbook is different from mine. Hence I would like you to verify my solution, and point out where I have made the mistake.

My take on the problem:

Let A be angle of the inclined plane

Due to gravity, acceleration of the object (assuming it is in free fall) equals to $32$ ft/s

If we are to include the fact that object slides downwards on an inclined plane, then acceleration will equal to:

$$a = -32\sin A$$

Since $a = \frac{dv}{dt}$, $v = \int -32\sin A dt = -32t \cdot \sin A +C$

Object starts at rest, hence $C = 0$, and thus $v = -32t \cdot \sin A$

Since $v = \frac{ds}{dt}$, we have $$s = -16t^2 \cdot \sin A + C$$

Object starts moving from the height h, then $C = h$, and therefore:

$$s = -16t^2 \cdot \sin A + h$$

We are asked to find time when the object reaches the ground. To do so, we let $s = 0$: $$-16t^2 \cdot \sin A + h = 0 \implies $$ $$16t^2 \cdot \sin A = h \implies $$ $$16t^2 = \frac{h}{\sin A} \implies $$ $$16t^2 = \frac{h}{\frac{h}{l}} = \frac{lh}{h} = l \implies $$ $$t^2 = \frac{l}{16} \implies $$ $$t = ±\frac{\sqrt{l}}{4}$$

We only consider positive t, hence: $$t = \frac{\sqrt{l}}{4}$$

However, the answer given by the textbook is $\frac{l}{4\sqrt{h}}$. Unfortunately, books does not provide explanation at how the solution was obtained.

Where have I made the mistake?

Answer 1

There is an error in your calculation. The acceleration $32\sin A$ is along the slanted plane, not vertically. Thus, the object travels a distance of $l$, not $h$.

So, the correct equation after your derivation is instead,

$$-16t^2\sin A + l = 0,$$

which leads to the correct answer.

Answer 2

I think your mistake is in deriving the constant of integration $C$.

$s$ measures distance along the plane, not vertical distance. So you should have $C=l$, not $C=h$.

If the acceleration due to gravity is $g$ then you should get

$t=l\sqrt{\frac{2}{gh}}$

If $l$ is in feet and you take $g$ to be $32$ feet per second$^2$ then you have

$t=\frac{l}{4\sqrt{h}}$ seconds

Answer 3

By the physics rules, I have that: $l=\frac{1}{2}at^2$. Substituing: $l=\frac{1}{2}g\cdot \sin(\alpha)\cdot t^2$. Also: $\sin(\alpha)=\frac{h}{l}$. From this: $2l^2=ght^2$ and so: $t=l\sqrt{\frac{2}{hg}}$.