Question

A measuring tape drops from a roof that is 16 feet above the ground. After how many seconds does the measuring tape land on the ground? How would you write the equation?

Answer 1

Answer:

It takes 1 second for the tape to reach the ground.

Equation to use: $y(t)=16-\frac{1}{2} g\,t^2[/tex with acceleration due to gravity "g" = 32ft/s^2]Step-by-step explanation:This is an object moving vertically under the action of the acceleration of gravity (32 ft/s^2), with a starting position of 16 feet, and with NO initial velocity (drops from the roof).The equation that describes the"y" position of the object as a function of time (t) will be written as:[tex]y(t) = y_0+ v_0*t-\frac{1}{2} g\,t^2$

As explain above, the initial position $y_0$ s 16 ft, there is no initial velocity, so $v_0=0$, 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:

$y(t) = y_0+ v_0*t-\frac{1}{2} g\,t^2\\y(t)=16-\frac{1}{2} g\,t^2$

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):

$y(t)=16-\frac{1}{2} g\,t^2\\0=16-\frac{1}{2} g\,t^2\\-16=-\frac{1}{2} 32\,t^2\\-16=-16\,t^2\\t^2=1\\t=+/- 1 \,second$

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.