A ball is thrown vertically upwards with an initial velocity of 27.0 m/s. Neglecting air resistance, how long is the ball in the air?
When h = hmax, Vf = 0,
Vf = Vi + gt = 0,
27 - 9.8t = 0,
-9.8t = -27,
t(up) = -27 / -9.8 = 2.76s,
t(tot.) = 2 * 2.76 = 5.52s.
To determine how long the ball is in the air, we need to find the time it takes for the ball to reach its maximum height and then double that time.
The initial velocity when the ball is thrown upwards is 27.0 m/s. We know that the final velocity at the highest point will be zero as the ball momentarily stops before falling back down due to gravity. We can use the following equation to determine the time taken to reach the maximum height:
vf = vi + at
where:
vf = final velocity (zero in this case)
vi = initial velocity (27.0 m/s)
a = acceleration (due to gravity, -9.8 m/s^2)
t = time
Rearranging the equation to solve for time, we have:
t = (vf - vi) / a
Since vf is zero, the equation becomes:
t = -vi / a
Plugging in the values:
t = -27.0 m/s / -9.8 m/s^2
The negative signs cancel out, so:
t = 27.0 m/s / 9.8 m/s^2
Calculating this:
t ≈ 2.76 seconds (rounded to two decimal places)
Since the ball takes the same amount of time to reach its maximum height as it does to fall back down, we double the result:
Total time = 2 * 2.76 seconds
Total time ≈ 5.52 seconds (rounded to two decimal places)
Therefore, the ball is in the air for approximately 5.52 seconds.