Questions: A ball is thrown upward with a velocity of 19.6 m / s. What is its velocity after 3.00 s? Multiple Choice 19.6 down 9.80 m / s down 9.80 m / s up zero

Solution Steps

We need to determine the velocity of a ball thrown upward with an initial velocity of \(19.6 \, \text{m/s}\) after \(3.00 \, \text{s}\). The ball is subject to gravitational acceleration, which is approximately \(9.81 \, \text{m/s}^2\) downward.

The kinematic equation for velocity under constant acceleration is: \[ v = u + at \] where:

• \(v\) is the final velocity,
• \(u = 19.6 \, \text{m/s}\) is the initial velocity,
• \(a = -9.81 \, \text{m/s}^2\) is the acceleration due to gravity (negative because it is downward),
• \(t = 3.00 \, \text{s}\) is the time.

Substitute the known values into the equation: \[ v = 19.6 \, \text{m/s} + (-9.81 \, \text{m/s}^2) \times 3.00 \, \text{s} \] \[ v = 19.6 \, \text{m/s} - 29.43 \, \text{m/s} \] \[ v = -9.83 \, \text{m/s} \]

The negative sign indicates that the velocity is directed downward. Therefore, the velocity of the ball after \(3.00 \, \text{s}\) is \(9.83 \, \text{m/s}\) downward.

Final Answer