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?
Best Answer
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.