rationalskepticism.org

Here's the sort of problem I give to my students in their 'Introduction to Mechanics' section (a bit of GCSE/AS physics crossover). To solve the problem we use the bit of physics and maths introduced by Hack and Darkchilde.

"MrC foolishly drives his 1981 Volvo estate car off a cliff. The car is intially moving with only a horizontal speed of 10ms-1. The cliff is 50m high. At what speed will he meet his impending death?
(assume that there is no air resistance and take g=10ms-2)"

To solve this problem we need to split the motion of the car into two components; the horizontal and the vertical.
Forces in the vertical direction (gravity) will not affect motion in the horizontal (where we have assumed no forces).

First of all, the equations of motion can can written as;

1) [math]v=u+at *
2) [math]v^2=u^2+2aS
3) [math]S=ut+1/2at^2 *

where;

S=distance travelled
u=initial velocity
v=final velocity
a=acceleration
t=time

*note that equation 1) and 3) are just versions of the ones Darkchilde wrote.

Solving the vertical component of velocity

The car starts off with an initial vertical velocity of u=0ms-1, we also know that it is accelerating at a=10ms-2 and the distance it will fall is S=50m.

We want to find the cars final velocity as it hits the ground below, so we use:

[math]v^2=u^2+2aS

and root both sides to find [math]v;

[math]v=\sqrt{u^2+2aS}

then substitute the values for u,a and S;

[math]v=\sqrt{0^2+2.10.50}=\sqrt{1000}=32ms^{-1} (nearest whole number)

Combining Horizontal and Vertical components

So, we now have the vertical velocity at which the car hits the bottom.
Recall that we already now the horizontal velocity at which the car meets the ground, it will be 10ms-1 because horizontally we have ignored air resistance i.e. the car won't slow down / speed up in that direction.

To find the speed that the car hits, we need to combine the horizontal and vertical components using a bit of pythagorus. The resultant speed will be the hypotenuse of the right angled triangle with sides 10ms-1 and 32ms-1.



[math]v_{resultant}^2=v_{horizontal}^2+v_{vertcal}^2

[math]v_{resultant}=\sqrt{v_{horizontal}^2+v_{vertcal}^2}

[math]v_{resultant}=\sqrt{10^2+32^2}

[math]v_{resultant}=34ms^{-1}

So, I meet my death at a speed of 34ms-1



edit - for spelling! Also, I've been a bit 'cavalier' with the terms 'velocity' and 'speed'. To be clear; Velocity is the Speed in a stated direction.