A brass ball is shot vertically upward from the surface of an atmosphere-free planet with an initial speed of 20.0 m/s. One second later, the ball has an instantaneous velocity in the upward direction of 15.0 m/s.
-Determine the velocity of the ball when it returns to its original position. Note: assume the upward direction is positive.
A)+20 m/s
B)-20 m/s
C)+40 m/s
D)-40 m/s
E)zero m/s

Question

Quizplus · Accepted Answer

First, let's use the equation for instantaneous velocity:

v = u + at

where v is final velocity, u is initial velocity, a is acceleration, and t is time. We can use this to find the acceleration of the brass ball one second after it is shot upwards:

15 m/s = 20 m/s + a(1 s)

a = -5 m/s^2

The negative sign means that the acceleration is in the downward direction.

Next, we can use the kinematic equation for displacement:

s = ut + (1/2)at^2

where s is displacement. We want to find the displacement of the ball when it returns to its original position, which means that s = 0. We can rearrange the equation to solve for the time it takes for the ball to reach its maximum height:

s = ut + (1/2)at^2
0 = 20 m/s(t) + (1/2)(-5 m/s^2)(t^2)
t = 4 s

Thus, it takes 4 seconds for the ball to reach its maximum height and return to its original position. We can use the equation for final velocity to find the velocity of the ball when it returns:

v = u + at
v = 20 m/s + (-5 m/s^2)(4 s)
v = -20 m/s

The negative sign means that the velocity of the ball when it returns is in the downward direction, so the best choice is B, -20 m/s.

A Brass Ball Is Shot Vertically Upward from the Surface