Question

A bobsled slides down an ice track starting
(at zero initial speed) from the top of a(n)
198 m high hill.
The acceleration of gravity is 9.8 m/s2 .
Neglect friction and air resistance and de-
termine the bobsled’s speed at the bottom of
the hill.
Answer in units of m/s

Answer 1

is the sled on a slope, or are we assuming it's sliding straight down?

Answer 2

Well it doesn't give an angle or anything so your assumption is as good as mine

Answer 3

ok in that case i would use the special case equation of freefall

\[v _{f}^2 = v _{i}^2 + 2 ad\]

vf is final velocity, vi is initial velocity, a is acceleration, and d is displacement

Answer 4

Well, although you gotta love magic formulas like @binarymimic has posted... I personally rather to avoid it, as I can't really remember them well and if I'm not familiar with them I'm doomed.
Instead, I'll try to explain the idea behind it. It's a lot longer, but I think it's worth it

Well at first some seemingly trivial stuff, what is acceleration?
Acceleration comes to describe the change of the velocity (speed) of an object.
It is defined in away so it has a value which when multiplied by the time passed (in some 'timeunits') would produce the change in the velocity.
Means, if have an X,Y axes system, and we treat X as 'time' in some 'timeunits' and Y as value, then if we describe the acceleration as a function in that system, then we can say that this function describes the slope of the curve of the velocity function.
Therefore, the acceleration function is the derivative of the velocity function.

Now what is velocity? velocity describes in a similar way to the acceleration, a change.
But this change is in the 'distance' an object has gone through.
In similar way, we could say that the velocity function is the derivative of the distance function.

Now... since we know the acceleration function, and that is described as a polynomial (with degree of 0):
\[
A(x) = 9.8 \quad\quad \text{| Acceleration}
\]
Then finding anti-derivatives is not too hard (especially for polynomials):
\[
V(x) = \int{A(x) dx} = 9.8x + Vi \quad\quad \text{| Velocity} \\
D(x) = \int{V(x) dx} = \cdot \frac{9.8}{2} \cdot x^2 + Vi \cdot x + Di \quad\quad \text{| Distance}
\]
\(D_i \) and \( V_i \) are Initial distance and Initial velocity (at x= 0). In our case they are both 0.
Now, short example, we can find at what 'time' (!) the distance has reached a given value.
\[
D(x) = 198 \\
\frac{9.8}{2} \cdot x^2= 198 \\
x = \pm\sqrt{ \frac{198 \cdot 2}{9.8} } \\
\]
But since negative time is meaningless to us then
\[
x \ge 0 \quad \implies \quad x =\sqrt{ \frac{198}{4.9} }
\]
Now to find velocity at that point:
\[
V \bigg( \sqrt{ \frac{198}{4.9} } \;\bigg) = 9.8 \cdot\sqrt{ \frac{198}{4.9} } = 62.29
\]
So speed should be \( 62.29_{m/s} \)
Well, that idea alone is enough imo, but to make it look less magical let's see how we get to the formula.

We can get the magic formula if we say:
\[
D_f = \text{Distance at the end} \\
V_f = \text{Velocity at the end} \\
A = \text{Acceleration} \\ \;\\ \;\\
D(t) = D_f \\
A \cdot \frac{t^2}{2} + Vi \cdot t + Di -D_f = 0 \\
\Delta D = D_f - D_i \quad \to \quad -\Delta D = D_i - D_f\\
\frac{A}{2} \cdot t^2 + Vi \cdot t - \Delta D = 0 \\
t = \frac{-V_i \pm \sqrt{V_i^2 - 4 \cdot \frac{A}{2} \cdot (-\Delta D) } }{2 \cdot \frac{A}{2}} \\
t = \frac{-V_i \pm \sqrt{V_i^2 + 2 \cdot A \cdot \Delta D } }{A} \\ \;\\\;\\
V_f = V(t) \\
V_f = A \cdot \Bigg [ \frac{-V_i \pm \sqrt{V_i^2 + 2 \cdot A \cdot \Delta D } }{A} \Bigg] + V_i \\
V_f \ = -V_i \pm \sqrt{V_i^2 + 2 \cdot A \cdot \Delta D } + V_i \\
V_f = \pm \sqrt{V_i^2 + 2 \cdot A \cdot \Delta D } \\
V_f^2 = V_i^2 + 2 \cdot A \cdot \Delta D
\]
Where, \(V_f\) is the velocity at the end, \(V_i\) is the initial velocity, \(A\) is the object's (constant) acceleration and \( \Delta D \) is the object's displacement.

Hope it's clear (and you managed to read it all...)