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

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

$u=9.81\times 4=39.24 m/s$

Let us take h be the height of building

$h=ut+\frac{-1}{2}gt^2$

$h=39.24\times 2.5-\frac{1}{2}\times 9.81\times 2.5^2$

h=67.45 m