Question

question 4 (1 point) a ball is thrown upward at a velocity of 19.6 m/s. what is its velocity after 3.00 s? 9.8 m/s upward 9.8 m/s downward zero 19.6 downward

Explanation:

Step1: Recall the kinematic equation for velocity

The kinematic equation for velocity when an object is in free - fall (or thrown vertically) is \(v = v_0+at\), where \(v_0\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. For an object thrown upward near the surface of the Earth, the acceleration due to gravity \(a=- 9.8\ m/s^{2}\) (negative because it acts in the downward direction, opposite to the initial upward velocity). The initial velocity \(v_0 = 19.6\ m/s\) and \(t = 3.00\ s\).

Step2: Substitute the values into the equation

Substitute \(v_0 = 19.6\ m/s\), \(a=-9.8\ m/s^{2}\) and \(t = 3.00\ s\) into the equation \(v=v_0 + at\).
\(v=19.6\ m/s+(-9.8\ m/s^{2})\times3.00\ s\)
First, calculate the product \((-9.8\ m/s^{2})\times3.00\ s=-29.4\ m/s\)
Then, \(v = 19.6\ m/s-29.4\ m/s=-9.8\ m/s\)
The negative sign indicates that the velocity is in the downward direction.

Answer:

B. 9.8 m/s downward