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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.

Respuesta :

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

[tex]u=9.81\times 4=39.24 m/s[/tex]

Let us take h be the height of building

[tex]h=ut+\frac{-1}{2}gt^2[/tex]

[tex]h=39.24\times 2.5-\frac{1}{2}\times 9.81\times 2.5^2[/tex]

h=67.45 m