
Velocity of a body
Hello All,
We are trying to work out how fast a body (human, who is ok now..) would need to be travelling to cover 19 meters. The person 90kg was travelling in a straight line when he hit a motor vehicle that turned in front of him, he travelled 19 meters and struck a solid object that stopped him(Headbang). I need to know how fast he was travelling.
Thanks

More information is required just to determine if the problem is solvable or not. For example, we need to know the trajectory of his flight. If that is not given, then we'll probably need the mass of the car, the car's velocities before and after the collision, as well as the person's velocity before the crash.
Also note that even if there is a set velocity for person travelling 19 metres, because the person strikes a solid object, there will not be a set velocity, only a minimum

Thanks Gusbob,
I know that the person (on a motorcycle) hit the car on the front corner as it was turning, which lead me to believe that the speed of the car was approx 5 km/hr. I hope this helps, sorry to sound so dumb...

Ok since the person was riding a motorcycle I'll assume he was flung over the handle bars and thrown forwards and a little upwards and travelled with constant velocity.
If you can provide the angle of trajectory I can actually give you a minimum velocity required to travel 19m. This is the equation if you're interested:
$\displaystyle s = v \cos \theta t = \frac{2V^2\sin \theta \cos \theta}{g} = \frac{V^2}{g} $
Note that V is the combined vertical and horizontal velocity. $\displaystyle V\cos\theta $ is the horizontal velocity.
There is still not enough information so far but I can give some equations that will help.
WARNING: I'm using average values for everything. This may differ significantly based on the unique scenario.
A car is roughly 1700kg. It probably won't be hampered too much after the crash so lets put the velocity at 3km/h after crash.
The median crash speed for motorcycle accidents is around 30km/h and the average mass of a motorcycle is about 200kg. I would assume the motrocycle completely stops after the crash.
Let the velocity of the person we're trying to find = V.
To summarise (m = mass, u = initial velocity v = final velocity)
Motorcycle (1) Person (2) Car (3)
m1 = 200kg u1 = 30km/h v1 = 0km/h
m2 = 90kg u1 = 30km/h v2 = V km/h
m3 = 1700kg u3 = 5km/h v3 = 3 km/h
According to the conservation of momentum
$\displaystyle m_1u_1 + m_2u_2 + m_3u_3 = m_1V_1 + m_2v_2 + m_3v_3 $
Substituting in the given values
$\displaystyle 6000 + 2700 8500 = 0  1697 + 90V $
$\displaystyle 90V = 1897 $
$\displaystyle V = 21.08 \, km/h $
Again $\displaystyle V\cos\theta $ is the horizontal velocity.

Gusbob, Thanks for that
It appears that the trajectory was close to 30 degrees.
The rider would have been 900 mm from the ground at impact and hit the stationary object at the end of the "flight" at a hight of 600 mm.
Does this help
Cheers

Yes, this helps a lot.
So using the formula I provided,
$\displaystyle s = \frac{V^2}{g} $
$\displaystyle 19 = \frac{V^2}{9.8} $
$\displaystyle V^2 = 186.2 $
$\displaystyle V = 13.6455... \approx 13.65 \, ms^{1} $
$\displaystyle 13.65 \, ms^{1} = 49.14 \, kmh^{1} $
This is the combined vertical and horizontal velocity.
The horizontal velocity alone is
$\displaystyle V \cos 30 = 42.56 \, kmh^{1} $ (2 decimal places)
This is the minimum velocity required to reach 19 metres given the information we have. Because he struck an object, there is reason to believe it was more than this.