A construction worker accidentally drops a brick from a high scaffold. a. What is the brick's velocity after 4.0 s? b. How far does the brick fall during this time?

The velocity after 4 seconds is 40 m/s and the height is 80 m.

From the equations of motion under gravity;

v = u + gt

V = final velocity of the brick

u = initial velocity of the brick

g = acceleration due to gravity = 10 ms-2

t = time taken

Note that, u = 0 m/s because the brick was dropped from a height.

v = 0 + 10(4)

v = 40 m/s

The height is obtained from;

h = ut + 1/2 gt^2

u = 0 m/s because the brick was dropped from a height.

h =  1/2 gt^2

h = 1/2 × 10 × (4)^2

h = 80 m

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A. 39.2 m/s

B. 78.4 m

Explanation:

Data obtained from the question include:

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

A. Determination of the brick's velocity.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = gt

v = 4 × 9.8

v = 39.2 m/s

Thus, the brick's velocity after 4 s is 39.2 m/s

B. Determination of how far the brick fall in 4 s.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

h = ½gt²

h = ½ × 9.8 × 4²

h = 4.9 × 16

h = 78.4 m

Thus, the brick fall 78.4 m during the time.

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pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y