Question

A stone is thrown vertically upward from ground level at t = 0. At t=2.50 s, it passes the top of a tall building, and 1.50 s later, it reaches its maximum height. What is the height of the tall building? We assume an answer in meters.

Answer 1

Answer:67.45 m

Step-by-step explanation:

Given

at t=2.5 s it passes the top of a tall building and after 1.5 s it reaches maximum height

let u is the initial velocity of stone

v=u+at

0=u-gt

`u=9.81* 4=39.24 m/s`

Let us take h be the height of building

`h=ut+(-1)/(2)gt^2`

`h=39.24* 2.5-(1)/(2)* 9.81* 2.5^2`

h=67.45 m