Question

A ball is thrown downward from the top of a building with an initial speed of 25 m/s. It strikes the ground after 2.0 s. How high is the building, assuming negligible air resistance

Answer 1

Answer: The height of the building is 69.6 m

Step-by-step explanation:

To calculate the height of the building, we use second equation of motion:

`s=ut+(1)/(2)at^2`

where,

s = height of the building = ?

u = initial velocity of the ball = 25 m/s

a = acceleration due to gravity =
`9.8m/s^2`

t = time taken = 2.0 sec

Putting values in above equation, we get:

`s=(25* 2.0)+(1)/(2)* 9.8* (2.0)^2\\\\s=69.6m`

Hence, the height of the building is 69.6 m

Answer 2

Use one of the equations of motion under constant acceleration:-

s = ut + 0.5at^2 where s = distance, u - initial velocity, a = acceleration ( in this case it is gravity = 9.81 m s^-2) and t = time.

here we have s = 25*2 + 0.5*9.81 * 2^2

= 69.62 meters answer