Search results
Results from the WOW.Com Content Network
This velocity is the asymptotic limiting value of the acceleration process, because the effective forces on the body balance each other more and more closely as the terminal velocity is approached. In this example, a speed of 50 % of terminal velocity is reached after only about 3 seconds, while it takes 8 seconds to reach 90 %, 15 seconds to ...
A rocket's required mass ratio as a function of effective exhaust velocity ratio. The classical rocket equation, or ideal rocket equation is a mathematical equation that describes the motion of vehicles that follow the basic principle of a rocket: a device that can apply acceleration to itself using thrust by expelling part of its mass with high velocity and can thereby move due to the ...
This speed is the asymptotic limiting value of the speed, and the forces acting on the body balance each other more and more closely as the terminal speed is approached. In this example, a speed of 50.0% of terminal speed is reached after only about 3 seconds, while it takes 8 seconds to reach 90%, 15 seconds to reach 99%, and so on.
Equation [3] involves the average velocity v + v 0 / 2 . Intuitively, the velocity increases linearly, so the average velocity multiplied by time is the distance traveled while increasing the velocity from v 0 to v, as can be illustrated graphically by plotting velocity against time as a straight line graph. Algebraically, it follows ...
The velocity maxima and minima do not necessarily occur when the crank makes a right angle with the rod. Counter-examples exist to disprove the statement "velocity maxima and minima only occur when the crank-rod angle is right angled".
When the velocity changes sign (at the maximum and minimum displacements), the magnitude of the force on the mass changes by twice the magnitude of the frictional force, because the spring force is continuous and the frictional force reverses direction with velocity. The jump in acceleration equals the force on the mass divided by the mass.
To find the angle giving the maximum height for a given speed calculate the derivative of the maximum height = / with respect to , that is = / which is zero when = / =. So the maximum height H m a x = v 2 2 g {\displaystyle H_{\mathrm {max} }={v^{2} \over 2g}} is obtained when the projectile is fired straight up.
The range and the maximum height of the projectile do not depend upon its mass. Hence range and maximum height are equal for all bodies that are thrown with the same velocity and direction. The horizontal range d of the projectile is the horizontal distance it has traveled when it returns to its initial height (=).