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A brick is released with no initial speed from the roof of a building and strikes the ground in 2.50 s, encountering no appreciable air drag. (a) how tall, in meters, is the building? (b) how fast is the brick moving just before it reaches the ground? (c) sketch graphs of this falling brickâs acceleration, velocity, and vertical position as functions of time.

Respuesta :

Catya
initial velocity = 0
time = 2.50 s
acceleration = gravity, 9.8 m/s^2

use the equations for linear motion.
s = ut + (1/2)at^2
v = u + at

A. displacement given u, a and t.
s = (0)(2.50) + (1/2)(9.8)(2.50)^2
s = 0 + 4.9(6.25)
s = 30.625 m
three significant figures..
s = 30.6 m

B. final velocity given u, a and t
v = 0 + 9.8(2.50)
v = 24.5 m/s

C. graphs
s vs. t
object starts at zero and displacement increases linearly to (2.5, 30.6) and stops. slope of line is average velocity, s/t
v vs. t
object starts at zero velocity and increases linearly to (2.5, 24.5) and stops. slope of line is acceleration, v/t.
a vs. t
acceleration is gravity, it's constant horizontal line at; y = 9.8. stops at (2.5,9.8)

Answer:

Part a)

y = 30.625 m

Part b)

v = 24.5 m/s

Explanation:

Part a)

As we know that brick will hit the floor after t = 2.50 s

so here we will have

[tex]y = v_i t + \frac{1}{2}at^2[/tex]

[tex]y = 0 + \frac{1}{2}(9.8)(2.50^2)[/tex]

[tex]y = 30.625 m[/tex]

Part b)

velocity of the brick just before it will strike the ground is given as

[tex]v_f = v_i + at[/tex]

[tex]v_f = 0 + (9.8)(2.5)[/tex]

[tex]v_f = 24.5 m/s[/tex]

Part c)

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