Stones are thrown horizontally with the same velocity from the tops of two different buildings. One stone lands twice as far from the base of the building from which it was thrown as does the other stone. Find the ratio of the height of the taller building to the height of the shorter building.

Correction:

hi/h2 = 4.9*(2t)^2/4.9*t^2
h1/h2 = 4.9*4*t^2/4.9*t^2 = 4.

h1/h2 = 4.9*(2t^2)/4.9*t^2

h1/h2 = 4.9*4*t^2/4.9t^2= 4.

Well, this is quite a mathematical question, but let's try to put a funny spin on it!

Okay, picture this: Two stones decided to have a throwing competition from their respective buildings. Let's call them Tall Tower and Shorty Shack.

Now, Tall Tower and Shorty Shack were feeling a bit competitive and decided to throw their stones horizontally with the same velocity. The stones, being little daredevils, land at different distances from the base of the buildings.

The first stone - let's call it Steve - likes to take his time and lands twice as far from the base of his building as the second stone - let's call her Sally. Maybe Steve enjoys the view a little too much, who knows?

Now, let's get to the height differences. The question asks for the ratio of the height of the taller building (Tall Tower) to the height of the shorter building (Shorty Shack).

Imagine Tall Tower as the giant skyscraper in town, and Shorty Shack as a cute little cottage.

Tall Tower's height is represented by "T" and Shorty Shack's height is represented by "S."

So, we need to find the ratio T:S.

From the information given, we can surmise that the time it takes for the stone to fall will be the same for both buildings, since they have the same initial velocity.

Now, picture the stones falling gracefully to the ground. Steve, the stone from Tall Tower, travels a longer distance horizontally, which means he takes more time to reach the ground.

Sally, the stone from Shorty Shack, lands closer to her building, covering a shorter distance horizontally and taking less time to make it to the ground.

Since the time taken is the same for both stones, we can say:

Distance Steve fell horizontally = 2 * Distance Sally fell horizontally.

"But Clown Bot, what does this have to do with the heights of the buildings?"

Great question!

Remember that the stones were thrown horizontally, so their initial vertical velocities are the same (horizontal velocity does not impact vertical velocity).

The only thing that affects the time of flight is the height of the buildings. So, we can consider:

Height Steve fell vertically = Height Sally fell vertically.

Now, let's find the ratio T:S based on the heights:

T/S = Distance Steve fell horizontally / Distance Sally fell horizontally

But we know that Distance Steve fell horizontally = 2 * Distance Sally fell horizontally.

Substituting that in:

T/S = 2 * Distance Sally fell horizontally / Distance Sally fell horizontally

Simplifying:

T/S = 2

So, the ratio of the height of the taller building (Tall Tower) to the height of the shorter building (Shorty Shack) is 2:1.

In other words, Tall Tower is twice as tall as Shorty Shack! Just like in a comedy duo, where one partner is always taller than the other.

To solve this problem, we can use the equations of motion. Let's assume the height of the shorter building is "h" and the height of the taller building is "2h".

When a stone is thrown horizontally, it only has horizontal motion, and its vertical motion is due to gravity. In the absence of air resistance, the horizontal component of velocity remains constant, while the vertical component experiences constant acceleration due to gravity.

Let's denote the velocity of the stones as "v." Since the stones are thrown horizontally, the initial vertical velocity is zero.

The time taken for a stone to reach the ground can be found using the equation:

h = (1/2) * g * t^2 --(1)

where g is the acceleration due to gravity (9.8 m/s^2).

The horizontal distance traveled by the stone can be calculated using the equation:

d = v * t --(2)

where d is the horizontal distance and t is the time.

In the problem, it is given that one stone lands twice as far from the base of its building compared to the other stone. Thus, we have:

d_1 = 2 * d_2 --(3)

Substituting equation (2) into (3) gives:

v * t_1 = 2 * v * t_2 --(4)

Cancelling out 'v' from both sides of equation (4) gives:

t_1 = 2 * t_2 --(5)

Now, let's solve equations (1) and (2) for the shorter building (h) and the taller building (2h).

For the shorter building:
From equation (1):
h = (1/2) * g * t_1^2 --(6)

From equation (2):
d_1 = v * t_1 --(7)

For the taller building:
From equation (1):
2h = (1/2) * g * t_2^2 --(8)

From equation (2):
d_2 = v * t_2 --(9)

Now, we can substitute equation (5) into equations (6) and (8) to get:

h = (1/2) * g * (2 * t_2)^2 --(10)

2h = (1/2) * g * t_2^2 --(11)

Simplifying equation (10) gives:

h = 2 * g * t_2^2 --(12)

Dividing equation (11) by equation (12) eliminates g and t_2 from the equation:

(2h) / h = (1/2) * g * t_2^2 / (2 * g * t_2^2)

Simplifying further, we get:

2 = 1/4

Therefore, the ratio of the height of the taller building to the height of the shorter building is 2:1.

Δy=Voy*t+1/2*ay*t^2

Your looking for height so you use the Motion in Two Dimensions in the y direction. Voy is the initial velocity of zero time time (t). Since zero times anything is zero the first part of the equation is zero leaving +1/2*ay*t^2. a is a constant and can be substituted for -9.8 (approx rate of gravity. 1/2*-9.8 is -4.9 leaving -4.9*t^2. You do the same again from the building the rock landed further away but since the rock was rwice the distance this will mean twice the time so after subing in you will have -4.9*(2t)^2. To find the ratio you do -4.9*t^2/-4.9*(2t)^2 leaving you with a ratio of 4/1=4.