Motorcycle stunt-rider mechanics problem

[look416]

Homework Statement

A motorcycle stunt-rider moving horizontally takes off from a point 1.25 m above the ground,
landing 10 m away as shown.

Homework Equations

The Attempt at a Solution

using tan [tex]\alpha[/tex] = [tex]\frac{Opp}{Adj}[/tex]
[tex]\alpha[/tex] = 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 [tex]\alpha[/tex]
therefore x = [tex]\frac{5}{sin\alpha}[/tex]
x = 40 ms-1
><
no the answer is 20ms-1
what is wrong?

 

[Andrew Mason]

Homework Statement

A motorcycle stunt-rider moving horizontally takes off from a point 1.25 m above the ground,
landing 10 m away as shown.

It is always a good idea to tell us what the question is. I take it that it is asking for the speed of the motorcycle when it takes off.
...

Your equation is not correct. The correct equation is:

[tex]h = v_0\sin{\alpha}t - \frac{1}{2}gt^2[/tex]

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

 

[look416]

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 ..