1. The maximum running speed (S) in km/hr and the corresponding body mass (m) in kg were determined for 55 different species of mammals. An empirical relation was determined to be:
S[km/hr] = 5.7 (m[kg]). Note: The notation in this equation means, "inserting the mass (m) in units of
‘kg', automatically calculates the maximum running speed (S) in units of ‘km/hr.
(a) What are the implied units for the "5.7" factor? Hint: recall that relations between physical quantities
must be dimensionally consistent, and the chosen units must in turn be consistent with the
dimensionality.
(b) Re-write the equation so that it gives the maximum speed in units of “m/s" for a given body mass in
"g”. Show all relevant details. If done correctly, you will need to use at least one unit conversion
factor.

A) 5.7 = (S/m)(km/(kg•hr))  

B) S (m/s) = 5.7(3600m) g

This is about conversion of units.

We are told that the empirical relation is given by;

S[km/hr] = 5.7 (m[kg])

(a) S[km/hr] = 5.7 (m[kg])

Since we are told that;

S is maximum running speed in km/hr

m is mass in kg

Making 5.7 the subject of the formula, we have to divide both sides by (m[kg]) to get;

5.7 = S[km/hr]/(m[kg])

Rearranging gives;

5.7 = (S/m)(km/(kg•hr))

(b) S[km/hr] = 5.7 (m[kg])

We want convert km/hr to m/s.

From conversions, we know that;

36 km/hr = 10 m/s

Thus; S km/hr = 10S/36 m/s = S/3.6 m/s

Similarly, from conversion tables, from kg to g;

1 kg = 1000 g

Thus;

m kg = (1000m) g

Thus; S[km/hr] = 5.7 (m[kg]) can now be expressed as;

S/3.6 m/s = 5.7(1000m) g

Thus, rearranging we have;

S (m/s) = (5.7 × 3.6 × 1000m) g

S (m/s) = 5.7(3600m) g

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(a) [km/hr/kg]

(b) S[m/s] = (5.7 / 3600) (m[g])

Explanation:

(a) S[km/hr] = 5.7 (m[kg])

Solving for 5.7:

5.7 = S[km/hr] / (m[kg])

5.7 = S/m [km/hr/kg]

(b) S[km/hr] = 5.7 (m[kg])

Convert km/hr to m/s.

S[m/s] = S[km/hr] × [1000 m/km] × [1 hr / 3600 s]

S[m/s] = S[km/hr] / 3.6

3.6 S[m/s] = S[km/hr]

Convert kg to g.

m[g] = m[kg] × [1000 g/kg]

m[g] = 1000 m[kg]

m[g] / 1000 = m[kg]

Substitute.

3.6 S[m/s] = 5.7 (m[g] / 1000)

S[m/s] = (5.7 / 3600) (m[g])

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

the correct choice is a .

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