Question

8. A stone that is thrown from the top of the "Main" building of the Polytechnic University is given an initial speed of 20.0 m / s straight up. The building is 50 meters high and the stone barely pounds the edge of the roof on its way down, as shown in the figure.

a) Use to = 0 as the time when the stone leaves the pitcher's hand in position and determine the time in which the stone reaches its maximum height.

b) Find the maximum height of the stone.

c) Determine the speed of the stone when it returns to the height from which it was thrown.

Answer 1

a) At maximum height, the speed will be zero,

so,

v = u + at

where

v - final velocity, u is initial velocity, a is acceleration

0 = u - gt

u = gt

t = u / g

t = 20 / 9.8

t = 2.0408 seconds

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b) v² - u² = 2gh

using same concept as (a) ,, set v = 0

h = u² / 2g

h = 20² / 2 * 9.8

h = 20.4 m

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c) v² - u² = 2gh

here , we set u = 0 because it starts from highest point and is moving down

so,

v = sqrt ( 2gh)

v = 20 m/s

as we can see it is same speed with which it was launched.

however, we need to use a negative sign to denote downward velocity

so,

v = - 20 m/s