How long does a ball stay in the air if it is 20m/s and it is thrown straightup?

V = Vo + g*Tr = 0

20 + (-9.8)Tr = 0
Tr = 2.04 s. = Rise time.
Tf = Tr = 2.04 s. = Fall time.
T = Tr+Tf = 2.04 + 2.04 = 4.08 s. = Time in air.

To determine how long a ball stays in the air when thrown straight up with an initial velocity of 20 m/s, you need to calculate the time it takes for the ball to reach the peak of its trajectory and then to return back to the ground.

The first step is to find the time it takes for the ball to reach its highest point (when its vertical velocity becomes zero). In this case, the initial velocity is positive (upward), and the acceleration due to gravity is negative (downward). To find the time, you can use the kinematic equation:

vf = vi + at

Since the final velocity (vf) at the highest point is zero, and the initial velocity (vi) is 20 m/s, you can set up the equation as follows:

0 = 20 m/s - 9.8 m/s^2 * t

Simplifying the equation gives you:

9.8 m/s^2 * t = 20 m/s

t = 20 m/s / 9.8 m/s^2

t ≈ 2.04 seconds

So, it takes approximately 2.04 seconds for the ball to reach its peak.

To find the total time the ball stays in the air, you need to double the time it took to reach the peak since it takes the same amount of time to descend. Therefore, the total time the ball stays in the air is roughly 2.04 seconds * 2 = 4.08 seconds.

Hence, the ball would stay in the air for approximately 4.08 seconds if it is thrown straight up with an initial velocity of 20 m/s.

To calculate the total time a ball stays in the air after being thrown straight up, we need to consider the motion of the ball.

When the ball is thrown straight up, it will experience a constant acceleration due to gravity, which is approximately 9.8 m/s². Since the initial velocity is 20 m/s and it is thrown straight up, the final velocity at the highest point will be zero.

We can use the kinematic equation to determine the time it takes for the ball to reach its highest point. The equation is:

v = u + at,

where:
v is the final velocity (0 m/s),
u is the initial velocity (20 m/s),
a is the acceleration (-9.8 m/s²), and
t is the time.

Plugging in the given values, we can calculate the time it takes for the ball to reach its highest point:

0 = 20 + (-9.8)t,
-9.8t = -20,
t = 2.04 seconds (rounded to two decimal places).

Therefore, it takes approximately 2.04 seconds for the ball to reach its highest point. To get the total time the ball stays in the air, we double this time:

Total time = 2 × t = 2 × 2.04 = 4.08 seconds (rounded to two decimal places).

So, the ball stays in the air for approximately 4.08 seconds.