Brad tosses a rock straight up at 38.4 m/s. What is the rock's velocity after 2.0 seconds?
V = 38.4 - g t
In this case, g = 9.8 m/s^2 and t = 2.0 s
153.68m/s^2
To determine the rock's velocity after 2.0 seconds, we need to consider the effects of gravity.
When an object is thrown straight up, its velocity decreases due to the acceleration from gravity. The acceleration due to gravity on Earth is approximately 9.8 m/s^2.
To find the final velocity, we can use the following equation:
v_f = v_i + (a * t)
Where:
v_f = final velocity
v_i = initial velocity
a = acceleration
t = time
Given:
v_i = 38.4 m/s (initial velocity)
t = 2.0 s (time)
a = -9.8 m/s^2 (acceleration due to gravity, negative because it acts opposite to the initial velocity)
Substituting the given values into the equation:
v_f = 38.4 m/s + (-9.8 m/s^2 * 2.0 s)
Simplifying the equation:
v_f = 38.4 m/s - 19.6 m/s
Therefore, the rock's velocity after 2.0 seconds is:
v_f = 18.8 m/s