How many feet tall is the building

1 answer:

6 0

Answer:

Height of the building = 77 ft.

Step-by-step explanation:

Let the height of the building (AB) = h feet

It's given that height of a woman (ED) = 5 feet 3 inches

≈ 5 feet + $\frac{3}{12}$ ft

≈ 5.25 ft

Length of the shadow (BC) of the building = 55 ft

Length of the shadow (DC) of the woman = 45 in ≈ 3.75 ft

Since, both the triangles ABC and EDC are similar, their corresponding sides will be proportional.

By the property of similar triangles,

$\frac{AB}{ED}= \frac{BC}{DC}$

$\frac{h}{5.25}=\frac{55}{3.75}$

h = $\frac{55\times 5.25}{3.75}$

h = 77 ft

Therefore, height of the building will be 77 ft.

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OlgaM077 [116]

Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

For a constant acceleration: $v_{f}=v_{0}+at$ , where $v_{f}$ is the final velocity in a direction after the acceleration is applied, $v_{0}$ is the initial velocity in that direction before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
Then for the x direction it is known that the initial velocity is $v_{0x} =$ 5320 m/s, the acceleration (the applied by the engine) in x direction is $a_{x}$ 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the is 739 s. Then: $v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}$
In the same fashion, for the y direction, the initial velocity is $v_{0y} =$ 0 m/s, the acceleration in y direction is $a_{y}$ 7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$