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geniusboy [140]
2 years ago
7

A rock is dropped from the top of a building and hits the ground at a velocity of

Mathematics
2 answers:
soldier1979 [14.2K]2 years ago
8 0

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

<h3>What is the equation of motion?</h3>

The equation of a motion is:

v^{2} =u^{2} +2gh

Where u and v are the initial and final velocities.

72^{2} =0+2*32*h\\144 = 64h\\h=81

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:

brainly.com/question/453639

Makovka662 [10]2 years ago
7 0

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:

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

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

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

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