A rock dropped from the top of a building takes 0.55 s to fall the last 50% of the distance from the top to the ground.How tall is the building?

Bob Pursley Said:

**hf=hi/2 + v'*.55+1/2 g .55^2
but v' is the velocity after it fell the first half
v'=g* timefirsthalf and
hi/2=1/2 g timefirsthalf^2
or timefirsthalf=sqrt (hi/g)
putting this into the equation **,
0=hi/2+.55(g*sqrt(hi/g))+4.86*.55*2

and set hf to zero but we are adding hi/2 + h(second half) and equation should be:
hi=hi/2+.55(g*sqrt(hi/g))+4.905*.55^2
and when you solve for x you get x = 17.2960032

17

Well, I don't want to rock your world, but let's calculate this together. So, if it takes 0.55 seconds for the rock to fall the last 50% of the distance, we can assume that it takes another 0.55 seconds for the rock to fall the first 50% of the distance.

Since the time it takes to fall from the top to the ground is the sum of these two times, we can say that it takes a total of 1.1 seconds for the rock to fall from the top to the ground.

Now, we need to use a bit of physics to solve this. The distance an object falls is given by the equation d = 1/2 * g * t^2, where d is the distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.

We can plug in the value of t as 1.1 seconds into this equation and since we want to find the total height of the building, we can simply multiply the result by 2 (as the rock only fell 50% of the distance).

So, using some calculations and mathematical magic, we find that the height of the building is approximately 6.5 meters. Hope this answer rocks your world!

To determine the height of the building, we need to understand the concept of free fall motion.

When an object is dropped, it falls freely under the force of gravity. The distance covered during free fall can be calculated using the equations of motion. In this case, we are given that it takes 0.55 s for the rock to fall the last 50% of the distance from the top to the ground.

Let's break down the problem step by step:

1. Suppose the total distance from the top of the building to the ground is represented by "d".
2. We are given that the rock takes 0.55 s to fall the last 50% of the distance (d/2) from the top to the ground.
3. During this time, the rock covers half the distance from the top to the ground. Therefore, the time taken to cover the remaining half of the distance is also 0.55 s.
4. Since the rock experiences uniform acceleration due to gravity, we can use the equation of motion: d = (1/2)gt^2, where d is the distance, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
5. Plugging in the values, we have d/2 = (1/2)(9.8)(0.55)^2.

Now, we can solve for "d" to find the height of the building:

d/2 = (1/2)(9.8)(0.55)^2
d/2 = 1.994
d = 2 * 1.994
d = 3.988 m

Therefore, the height of the building is approximately 3.988 meters.

idont understand

let h be the height of the building

and hf be the base, or 0

**hf=hi/2 + v'*.55+1/2 g .55^2
but v' is the velocity after it fell the first half
v'=g* timefirsthalf and
hi/2=1/2 g timefirsthalf^2
or timefirsthalf=sqrt (hi/g)
putting this into the equation **,
0=hi/2+.55(g*sqrt(hi/g))+4.86*.55*2

solve for hi. It appears to be a quadratic, I think to simplify it I would substutute x=sqrt(hi) and solve for x with the quadratic equation, then find hi.