# Calculate the velocity and the time for a ball to strike the ground if the ball is dropped from the top of 40 m tall building

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### 2 Answers

For a ball dropped from the top of a building the acceleration downwards due to gravity is 9.8 m/s^2.

Now the height from which it is dropped is 40m. And as it is dropped its initial velocity was 0 m/s^2.

Now let’s say the ball strikes the ground after a time t. The velocity after t sec is v= 0 + 9.8* t

The distance travelled in time t is the product of the average velocity and the time.

So 40 = (0 + 9.8t)*t/2

=> 9.8t^2 = 80

=> t = sqrt [80/9.8] = approximately 2.85 sec

So the ball takes 2.85 s to strike the ground and its velocity is 9.8*2.85 = 28 m/s.

**The required values are: time taken 2.85s and velocity 28 m/s**

Since to calculate velocity, we need to know the time, we'll determine the time first.

To calculate the time, we'll use the equation:

x - x0 = v0*t + a*t^2/2 (1)

x - x0 is the distance covered by the ball when it's dropped from the top of the building.

Since the building is 40 m tall, then the distance is the height of the building.

x - x0 = 40m

The original velocity is 0, v0 = 0.

The acceleration is g = 9.8 m/s^2

We'll substitute all these values in (1):

40 = 0 + 9.8*t^2/2

80 = 9.8*t^2

We'll divide by 9.8 and we'll get the time:

t^2 = 80/9.8

t = 2.85 s approx.

round t

**t = 2.9 s**

Now, we can calculate the velocity, because we've get the time:

v = v0 + g*t

v = g*t

v = 9.8*2.9

**v = 28 m/s**