Chapter 6.1, Problem 11SSC

To determine

The effect of the acceleration on the parameters of the motion.

Answer to Problem 11SSC

The vertical velocity of the object remains unchanged on the moon and time of flight, maximum height and horizontal distance of the object on the moon increases.

Explanation of Solution

The vertical velocity of the object is given as,

  $v_y=v\sin \theta$

The time of flight is given as,

  $t=\frac{v\sin \theta }{g}$

The maximum height of the object is given as,

  $H=\frac{v^2\sin 2\theta }{g}$

The horizontal distance of the of the object is given as,

  $R=\frac{v^2{\sin }^2\theta }{g}$

The vertical velocity does not depend on the acceleration, so the vertical velocity of the object remain unchanged on the moon.

The time of flight, maximum height and horizontal distance of the object depend on acceleration with inverse relation. The acceleration on the moon becomes one sixth of the acceleration on the earth, so the time of flight, maximum height and horizontal distance of the object become six time to the corresponding values on the earth.

Thus, the vertical velocity of the object remain unchanged on the moon and time of flight, maximum height and horizontal distance of the object on the moon increases.