Question

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

Answer 1

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

What is the equation of motion?

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.

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Answer 2

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:

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