Motorcycle stunt-rider mechanics problem

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SUMMARY

The discussion focuses on solving a physics problem involving a motorcycle stunt-rider who takes off from a height of 1.25 meters and lands 10 meters away. The correct approach to determine the motorcycle's speed at takeoff involves using the equation h = v_0 sin(α)t - (1/2)gt², where h represents the vertical change, g is the acceleration due to gravity (10 m/s²), and α is the launch angle. The angle of launch is established as 0 degrees, leading to the calculation of time of flight to find the horizontal speed, which is ultimately determined to be 20 m/s.

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  • Familiarity with kinematic equations
  • Knowledge of trigonometric functions
  • Basic grasp of gravitational acceleration (g = 10 m/s²)
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look416
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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.

Untitled.png


Homework Equations





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
><
no the answer is 20ms-1
what is wrong?
 
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look416 said:

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.

Untitled.png
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:

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
 


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 ..
 

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