A Ski jumper starts at point A and leaves at point B. Assume a(naught)=gsin(theta) tangent to path, where g=9.81m/s^2. Determine takeoff speed, v, if r=60m and theta=40 degrees. (Hint: You will need the chain rule.)
To determine the takeoff speed of the ski jumper, we need to use the given information about the path of the jumper and apply the chain rule to find the derivative of the displacement with respect to time.
Let's denote the takeoff speed as v and the time as t. Since the ski jumper starts at point A and leaves at point B, we can consider the horizontal distance traveled as displacement. Let's call this displacement r.
Now, we can express r as a function of time (r = r(t)). Since we know the initial velocity in the horizontal direction is zero (naught), the only force acting on the ski jumper in the horizontal direction is the component of the gravitational force parallel to the path, which is given as naught = g * sin(theta).
Using the equation of motion in the horizontal direction, we have:
r = naught * t
Since naught is a constant, we can differentiate both sides of the equation with respect to time:
dr/dt = d(naught * t)/dt
The derivative of naught with respect to time is zero, so we have:
dr/dt = naught
Now, we can substitute the expression for naught:
dr/dt = g * sin(theta)
This equation tells us that the rate of change of the displacement with respect to time is the horizontal component of the gravitational force, which is g * sin(theta).
Now, let's use the chain rule to find the derivative of r with respect to t on the left side of the equation:
dr/dt = (dr/dv) * (dv/dt)
The derivative of r with respect to v (dr/dv) is simply the inverse of the derivative of v with respect to r, so we can express it as:
dr/dv = 1 / (dv/dr)
Now, we need to differentiate the equation r = naught * t with respect to r:
dr/dv = 1 / (dv/dr) = 1 / (d(naught * t)/dr)
Since naught is a constant, we can treat it as such when differentiating:
dr/dv = 1 / (naught * dt/dr)
Now, we can solve for dt/dr using the equation r = naught * t:
r = naught * t
dt/dr = 1 / (naught * dr/dt)
dt/dr = 1 / (g * sin(theta))
Substituting this expression back into the equation for dr/dv:
dr/dv = 1 / (naught * dt/dr) = 1 / (naught * (1 / (g * sin(theta))))
Now, we can substitute this expression for dr/dv into the first equation:
g * sin(theta) = (dr/dv) * (dv/dt)
Since dr/dv = 1 / (naught * (1 / (g * sin(theta)))) and dv/dt = v, we can simplify the equation as:
g * sin(theta) = (1 / (naught * (1 / (g * sin(theta))))) * v
Simplifying further:
v = g * sin(theta) * naught^2 / (g * sin(theta))
Finally, we can substitute the given values, g = 9.81 m/s^2, theta = 40 degrees, and r = 60 m:
v = (9.81 m/s^2) * sin(40 degrees) * (naught^2) / (9.81 m/s^2 * sin(40 degrees))
Simplifying:
v = naught^2
Therefore, to find the takeoff speed v, we need to square the initial velocity naught, which is given as g * sin(theta). Given the values of theta and g, we can substitute them into the equation and calculate the takeoff speed.