A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude of 9 m/s. Find the motorcycle's position from the edge of the cliff afyer 0.5 s. Use g = 9.8 m/s.s
horizontal=9*.5=4.5m
vertical=1/2 g t^2=4.8(1/4)
then add these two displacements to give the vector distance.
Position r = 4.5 i + (-4.9*.5^2) j
To find position use theta=tan^-1(Vy/Vx)
To solve this problem, we can use the kinematic equations of motion. In this case, since we are given the initial velocity and time, and we need to find the position, we can use the equation:
y = y_0 + v_0y * t + (1/2) * g * t^2
Where:
- y is the position from the edge of the cliff after 0.5 s (which is what we want to find)
- y_0 is the initial position (which is the edge of the cliff, so it is 0)
- v_0y is the vertical component of the initial velocity (which is 0 since the velocity is horizontal at the edge of the cliff)
- t is the time (which is given as 0.5 s)
- g is the acceleration due to gravity (which is given as 9.8 m/s^2)
Plugging in the values into the equation, we have:
y = 0 + 0 * 0.5 + (1/2) * 9.8 * (0.5)^2
Simplifying the equation:
y = 0 + 0 + (1/2) * 9.8 * 0.25
y = 0 + 0 + 1.225
y = 1.225
Therefore, the motorcycle's position from the edge of the cliff after 0.5 s is 1.225 meters.