# Motorcycle stunt-rider mechanics problem

1. May 17, 2010

### look416

1. The problem statement, all variables and given/known data
A motorcycle stunt-rider moving horizontally takes off from a point 1.25 m above the ground,
landing 10 m away as shown.

2. Relevant equations

3. The attempt at a solution
using tan $$\alpha$$ = $$\frac{Opp}{Adj}$$
$$\alpha$$ = 7.1
then using v2=u2+2as
v2=0
let a = g = 10
s = 1.25(since we are calculating vertical)
u2 = 25
u = 5
but u is Xsin $$\alpha$$
therefore x = $$\frac{5}{sin\alpha}$$
x = 40 ms-1
><
what is wrong?

2. May 17, 2010

### Andrew Mason

Re: mechanics

Your equation is not correct. The correct equation is:

$$h = v_0\sin{\alpha}t - \frac{1}{2}gt^2$$

where h = change in vertical position = -1.25 m.

The angle of launch is 0 degrees. Work out the time of flight. From that you can work out its horizontal speed (it travels 10 m in that time).

AM

3. May 18, 2010

### look416

Re: mechanics

yea
i have found it
thx bro for stating the angle of launch
i always thinking that for no angle is given
we have to find the angle by ourselves using trigo ..