Answer:
It takes 1 second for the tape to reach the ground.
Equation to use: [tex]y(t)=16-\frac{1}{2} g\,t^2[/tex with acceleration due to gravity "g" = 32ft/s^2]
Step-by-step explanation:
This is an object moving vertically under the action of the acceleration of gravity (32 ft/s^2), with a starting position of 16 feet, and with NO initial velocity (drops from the roof).
The equation that describes the"y" position of the object as a function of time (t) will be written as:
[tex]y(t) = y_0+ v_0*t-\frac{1}{2} g\,t^2[/tex]
As explain above, the initial position [tex]y_0[/tex] s 16 ft, there is no initial velocity, so [tex]v_0=0[/tex], and the acceleration of gravity is 32 ft/s^2, and should be considered negative [as pointing down in the y-direction], so the equation simplifies to:
[tex]y(t) = y_0+ v_0*t-\frac{1}{2} g\,t^2\\y(t)=16-\frac{1}{2} g\,t^2[/tex]
In order to find the tima it takes it to hit the ground, we simple solve the equation for t when y(t) = 0 (the tape has reached the ground (zero height in the y-direction):
[tex]y(t)=16-\frac{1}{2} g\,t^2\\0=16-\frac{1}{2} g\,t^2\\-16=-\frac{1}{2} 32\,t^2\\-16=-16\,t^2\\t^2=1\\t=+/- 1 \,second[/tex]
We select the positive time (+1 second) which is what makes physical sense, since a negative value in time would mean time before the tape was dropped.
So the answer is: It takes 1 second for the tape to reach the ground.
The time taken for the measuring tape dropping from a height of 16 ft to land on the ground is 1 s
1 ft = 0.305 m
Therefore,
16 ft = 16 × 0.305
16 ft = 4.88 m
h = ½gt²
4.88 = ½ × 9.8 × t²
4.88 = 4.9 × t²
Divide both side by 4.9
t² = 4.88 / 4.9
Take the square root of both side
t = √(4.88 / 4.9)
t = 1 s
Thus, it will take the measuring tape 1 s to land on the ground.
Learn more about motion under gravity:
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