# A ball is thrown vertically upwards with a speed of 27.3 m/s from a height of 2.0 m. How long does it take to reach its highest point? How long does the ball take to hit the ground after it reaches its highest point? The acceleration of gravity is 9.82 m/s^2. Consider the two parts of the ball's motion separately.

First, the ball is traveling upward. Since the initial velocity is directed up, and the gravitational acceleration is directed down, the ball will slow down and eventually stop. Then, it will start moving down. So, at the highest point, the ball's velocity is 0.

We can use this equation of motion to find the time it takes the ball to reach the highest point:

`v_f = v_0 + at`

`v_f = 0` as discussed above, `v_0 = 27.3 m/s` and `a =-9.82 m/s^2` (negative, because it is in the direction opposite to initial velocity.)

Plugging these values into the equation results in

0 = 27.3 - 9.82t

From here `t = 27.3/9.82= 2.78 s`

The time it takes the ball to reach the highest point is 2.78 seconds.

To find how long it will take the ball to hit the ground afterwards, we first need to find the height of the highest point. Since we already know the time it took the ball to get there, we can use the equation

`h = h_0 + v_0t+at^2/2`

The initial height is `h_0 = 2m` , so

`h = 2 + 27.3(2.78) - 9.82(2.78)^2/2`  = 39.9 meters.

So the balls falls down from the height of h = 39.9 meters with initial velocity 0. The time it will take the ball to hit the ground is determined by

`0 = h +at^2/2`

`0 = 39.9 - 9.82*t^2/2`

From here `t = sqrt((2*39.9)/9.82) = ` 2.85 seconds.

It will take the ball 2.85 seconds after it reaches the highest point to hit the ground.

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