You throw a ball straight up at 25 m/s. How many seconds elapse before it is traveling downward at 13 m/s?
Vf=Vi-1/2 g t^2
solve for t when Vf=-13
3.8
To find the time it takes for the ball to transition from an upward velocity of 25 m/s to a downward velocity of 13 m/s, we can use the equation of motion:
vf = vi + at
where:
vf = final velocity (13 m/s)
vi = initial velocity (25 m/s)
a = acceleration (due to gravity, approximately -9.8 m/s^2)
t = time
Rearranging the equation, we have:
t = (vf - vi) / a
Substituting the given values into the equation:
t = (13 m/s - 25 m/s) / -9.8 m/s^2
Simplifying the equation:
t = -12 m/s / -9.8 m/s^2
t ≈ 1.22 seconds
Therefore, it takes approximately 1.22 seconds for the ball to transition from an upward velocity of 25 m/s to a downward velocity of 13 m/s.
To determine the time it takes for the ball to start traveling downward at 13 m/s, we can use the equations of motion.
When the ball is thrown straight up, its initial velocity (u) is 25 m/s. The final velocity (v) of the ball when it starts traveling downward is -13 m/s (negative sign indicates the downward direction). The acceleration due to gravity (a) is approximately -9.8 m/s² (negative sign indicates the opposite direction to the initial velocity).
We need to find the time (t) it takes for the ball to reach the downward velocity. We can use the following equation of motion:
v = u + at
Rearranging the equation to solve for time (t):
t = (v - u) / a
Substituting the given values:
t = (-13 - 25) / (-9.8)
t = -38 / -9.8
t ≈ 3.88 seconds
Therefore, approximately 3.88 seconds elapse before the ball is traveling downward at 13 m/s.