a person watching through a window of an apartment sees a ball that rise vertically up then vertically down for a total time of0.5s.if the hieght of the window is 2m ,find the max hieght reached by ball above the window

Are you sure the time is only o.5 s.?

To find the maximum height reached by the ball above the window, we can use the concept of projectile motion. Since the ball rises vertically up and then vertically down, we only need to consider the upward motion.

The key here is to use the time of flight, which is the total time the ball is in the air. In this case, the total time of flight is given as 0.5 seconds.

The time of flight is divided into two parts - the time of ascent (when the ball is rising) and the time of descent (when the ball is falling). Since the ball rises and falls for an equal amount of time, each part would be equal to 0.5 seconds divided by 2, which is 0.25 seconds.

Now, we can use the formula for calculating the maximum height reached by an object in projectile motion:

Height = (initial velocity * time) - (0.5 * acceleration * time^2)

Since the ball only rises vertically, the initial velocity will be in the upward direction, and the acceleration will be the acceleration due to gravity (g = 9.8 m/s^2) pointing downwards.

Let's plug in the values:

Initial velocity = ?
Acceleration = -9.8 m/s^2 (negative because the acceleration is downwards)
Time = 0.25 seconds

We need to find the initial velocity, so let's rearrange the formula:

Height + (0.5 * acceleration * time^2) = initial velocity * time

Substituting the given values:

2m + (0.5 * -9.8 m/s^2 * (0.25 s)^2) = initial velocity * 0.25s

Simplifying the equation:

2m - 0.30625m = initial velocity * 0.25s

1.69375m = initial velocity * 0.25s

initial velocity = 1.69375m / 0.25s ≈ 6.775 m/s

Therefore, the maximum height reached by the ball above the window is calculated by using this initial velocity in the formula for the height:

Height = (initial velocity * time) - (0.5 * acceleration * time^2)
Height = (6.775 m/s * 0.25 s) - (0.5 * -9.8 m/s^2 * (0.25 s)^2)
Height ≈ 1.6875m

So, the maximum height reached by the ball above the window is approximately 1.6875 meters.