Final answer:
It will take approximately 2.04 seconds for a ball thrown upward with an initial velocity of 20.0 m/s to reach its maximum height, calculated using the kinematic equation v = v_i + at with gravity acting as the acceleration.
Step-by-step explanation:
The question asks how long a ball thrown upward with an initial velocity of 20.0 m/s will take to reach its maximum height. To solve this, we'll use the kinematic equation for uniformly accelerated motion, where acceleration is due to gravity, which is approximately 9.81 m/s2 downward.
Firstly, we know that at the maximum height, the velocity of the ball will be 0 m/s. The initial velocity (vi) is 20 m/s upward, and the acceleration (a) due to gravity is -9.81 m/s2 (the negative sign indicates that it is acting in the opposite direction to the velocity).
The formula to calculate the time (t) to reach the maximum height is:
v = v
i
+ at
Where:
- v = final velocity (0 m/s at the maximum height)
- vi = initial velocity (20.0 m/s upward)
- a = acceleration due to gravity (-9.81 m/s2)
- t = time to reach the maximum height
By substituting the known values into the equation, we get:
0 m/s = 20.0 m/s + (-9.81 m/s2) * t
When we solve for t, we find:
t = (0 m/s - 20.0 m/s) / -9.81 m/s2
t = 20.0 m/s / 9.81 m/s2
t ≈ 2.04 seconds
Therefore, it will take approximately 2.04 seconds for the ball to reach its maximum height.