Abrick is thrown vertically upward with an initial speed of 3.00 m/s from the roof of a building. if the building is 78.4 m tall, how much time passes before the brick lands on the ground?

4.32 seconds

Explanation:

Given:

y = 0 m

y₀ = 78.4 m

v₀ = 3.00 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 = 78.4 + 3.00 t − 4.9 t²

4.9t² − 3t − 78.4 = 0

Solve with quadratic formula:

t = [ 3 ± √(9 − 4(4.9)(-78.4)) ] / 9.8

t ≈ -3.71, 4.32

Since t must be positive, t = 4.32.  So it takes 4.32 seconds for the brick to land.

trig 15.3 cos

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.