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A brick is thrown upward from the top of a building at an angle of 30° to the horizontal and with an initial speed of 14 m/s. if the brick is in flight for 3.1 s, how tall is the building?

Respuesta :

DeanR
[tex]v = 14 \textrm{ meters / second} \qquad \theta=30^\circ[/tex]

We don't care about the component of motion in the x direction.  Let's focus on the y direction.

[tex]v_y = v \sin \theta - gt[/tex]

where [tex]g=9.8 \textrm {m/s}^2[/tex] is the acceleration due to gravity.

Integrating, we have

[tex]y = vt \sin \theta - \frac 1 2 gt^2 + h[/tex]

[tex]h[/tex] is the arbitrary integration constant; let's let it be the unknown height of the building.

We hit the ground in 3.1 seconds:

[tex]0 = 14 (3.1) \sin 30^\circ - \frac 1 2 (9.8) (3.1)^2 + h[/tex]

[tex]h =\frac 1 2 (9.8) (3.1)^2 - 14 (3.1) (\frac 1 2) =25.389 \textrm{ meters}[/tex] 

The height of the building is 2.5 meters.

Given

A brick is thrown upward from the top of a building at an angle of 30° to the horizontal and with an initial speed of 14 m/s.

What is the formula of maximum height?

The formula is used to find maximum height is;

[tex]\rm H=\dfrac{u^2sin^2\theta}{2g}[/tex]

Substitute all the values in the formula

[tex]\rm H=\dfrac{u^2sin^2\theta}{2g}\\\\\rm H=\dfrac{(14)^2\times sin^2(30)}{2\times 9.8}\\\\H =\dfrac {196 \times 0.25}{19.6}\\\\H = \dfrac{49}{19.6}\\\\H = 2.5[/tex]

Hence, the height of the building is 2.5 meters.

To know more about Maximum height click the link given below.

https://brainly.com/question/5594783

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