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
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.
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