A rock is dropped from the top of a building and hits the ground at a velocity of
–72ft/sec. If the acceleration due to gravity is – 32ft /sec², what is the height of the
building?

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Answer :

Evgeniylevigo Evgeniylevigo · Answer 1

Answer:

81 [ft].

Step-by-step explanation:

1) the basic formula is: h=gt²/2, where g - acceleration due to gravity, t - elapsed time;

2) if the final velocity is 72, g=32, then it is possible to calculate elapsed time:

[tex]t=\frac{V}{g}=\frac{72}{32}=\frac{9}{4} [sec].[/tex]

3) if g=32, t=9/4, then the required height is:

[tex]h=\frac{32*(\frac{9}{4} )^{2} }{2}=\frac{16*81}{16}=81[ft].[/tex]

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AshishdwivedilVTgo AshishdwivedilVTgo · Answer 2

The height of the building comes to be 81 feet.

Initial velocity u= 0 feet/sec

Final velocity v= 72 feet/sec

The acceleration due to gravity g =32ft /sec²

Height of the building h= suppose h

The equation of a motion is:

[tex]v^{2} =u^{2} +2gh[/tex]

Where u and v are the initial and final velocities.

[tex]72^{2} =0+2*32*h\\144 = 64h\\h=81[/tex]

So, the height of the building = 81 feet.

Therefore, the height of the building comes to be 81 feet.

To get more about motion visit:

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