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For a body falling freely from rest (neglecting air resistance), the distance the body falls varies directly as the square of the time. If an object is dropped from a tower 400 ft high and hits the ground in 5 seconds, how far did it fall in the first 3 seconds?

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Hi, Daniel - I'm going to take a somewhat different approach than Vivian, using just the information given in the problem.

The problem states that the distance an object falls varies directly as the square of the time. Before we even go any further, let's use that information to develop an equation, involving the distance (d) that the object falls, and the time (t) that it takes to fall.

Because the distance (d) varies DIRECTLY (as opposed to inversely), we know that the distance (d) is a multiple of the time (t) squared:

d = something * t^{2}

We want to figure out exactly what that "something" is... I'm going to replace that "something" with the letter "A"

d = A * t^{2}

The other piece of information that the problem gives us is that the object falls 400ft in 5 seconds. Those are two corresponding (d) and (t) values, which, if we substitute them into our equation, should allow us to determine the specific value of A:

400 = A * 5^{2}

400 = A * 25

We can solve for A by dividing each side by 25:

400 = A * 25

25 25

16 = A

This allows us to explicitly define the relationship between the distance an object falls (d) and the time it takes (t), because we have used the given information to calculate that "something" that we used in our original equation - we now know that "something" equals 16.

d = 16 * t^{2}

Since we now know the specific equation that relates the distance fallen and time, we can plug in any value for time, and use the equation to calculate the distance an object falls in that amount of time. The question wants us to determine how far an object falls in 3 seconds, so let's plug in 3 for (t), and see what (d) value it gives us, using the known equation:

d = 16 * t^{2}

d = 16 * (3)^{2}

d = 16 * 9

d = 144 ft

I hope this helps!

Hi Daniel;

g=-9.81 m/s^{2}

This is the acceleration rate. The negative sign is because it is falling down.

s=(1/2)a(t^{2})

s=distance

a=acceleration

t=time

Our first priority is to convert feet into meters...

400 feet(1 meter/3.28 feet)

Let's cancel units...

Feet is in both the numerator and denominator...

400 feet(1 meter/3.28 feet)

(400 meters)/3.28

121.95 meters

Because the distance is from air to ground, this is negative...

-121.95 meters

s=(1/2)a(t^{2})

s=(1/2)(-9.81 m/s^{2})(3 s)^{2}

Let's cancel units.

s^{2} is in both the numerator and denominator...

s=(1/2)(-9.81 m)(3)^{2}

s=(1/2)(-9.81 m)(9)

s=-44.15 meters

Let's apply this equation to 5 seconds...

s=(1/2)a(t)^{2}

s=(1/2)(-9.81 m/s^{2})(5 s)^{2}

s=(1/2)(-9.81 m)(25)

s=-122.63 m

It is close to enough to -121.95 meters.

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