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

Answer:

h = 65.021 m

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 :

$h=ut+\dfrac{1}{2}at^2$

here, a = g

$h=17(2.3)+\dfrac{1}{2}\times 9.8\times 2.3^2\\\\=65.021\ m$

So, the building is 65.021 m tall.