a ball is thrown vertically upward at a speed of 45 m/s. how long does it take to return to the ground?

Tr = (V-Vo)/g = (0-45)/-9.8 = 4.59 s. =

Rise time.

Tf = Tr = 4.59 s. = Fall time.

Tr + Tf = 4.59 + 4.59 = 9.2 s. = Time to
return to Gnd.

To find out how long it takes for the ball to return to the ground, we need to determine the time it takes for the ball to reach its maximum height and then double that time.

First, let's analyze the motion of the ball. When the ball is thrown vertically upward, it experiences a constant acceleration due to gravity. The initial velocity is 45 m/s in the upward direction.

At the highest point of the ball's trajectory, its vertical velocity becomes zero. This means that the time taken to reach the maximum height is the same as the time taken to return to the ground.

We can use the equation of motion to calculate the time for the ball to reach its maximum height:

v = u + at

Where:
- v is the final velocity (zero at maximum height)
- u is the initial velocity (45 m/s upwards)
- a is the acceleration due to gravity (-9.8 m/s², negative because it acts in the opposite direction to the initial velocity)
- t is the time taken to reach the maximum height

Rearranging the equation, we get:

t = (v - u) / a

Substituting the values:

t = (0 - 45) / (-9.8)
t = 4.59 seconds (rounded to two decimal places)

Since the time taken to reach the maximum height is the same as the time taken to return to the ground, we double this time:

Total time taken = 4.59 * 2
Total time taken ≈ 9.18 seconds

Therefore, it takes approximately 9.18 seconds for the ball to return to the ground.