Question

A baseball is thrown off of a building with an initial velocity of 17 m/s and lands 2.3 seconds later. How tall is the building?

Answer 1

h = 65.021 m

Step-by-step explanation:

Given that,

Initial velocity of a baseball, u = 17 m/s

It lands 2.3 seconds later.

We need to find the height of the building. Let it is h. It can be calculated using second equation of motion as follows :

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

here, a = g

`h=17(2.3)+(1)/(2)* 9.8* 2.3^2\\\\=65.021\ m`

So, the building is 65.021 m tall.