physics help please?

Question

i tried everything i tried the pathagorians theorem and it didnt work. how on earth can i do this problem???

A snowball is launched horizontally from the top of a building at v = 11.4 m/s. If it lands d = 33 meters from the bottom, how high (in m) was the building?

Answer 1

Horizontal velocity is constant.
Hence time taken to reach the bottom = 33/11.4 s
If h = height of the building,
h = 4.9 t^2
= 4.9 * (33/11.4)^2
= 41 m.

Answer 2

Note: Pythagoras' Theorem has little to do with this problem.

First, let's assume that the horizontal velocity of the snowball does not change (since gravity only acts in the vertical direction.) Since velocity = distance / time, and since both velocity and horizontal distance are known constants, the time it takes the snowball to reach the ground is

t = x / v{x}

where v{x} is the horizontal velocity, and x is the distance the snowball lands from the building.

Now that we know the time of fall, we can use the equation

y = y{0} + v{0} t + a t² /2;

where y{0} is the initial position, v{0} is the initial velocity, and a is the acceleration of the object. (note that this equation is only valid if the acceleration is constant.)

If we consider that when the snowball splats on the ground, it's position is y = 0, that it has no initial downward velocity, and if we replace the acceleration with a = -g, the equation can be rewritten as

y{0} = g t² / 2.

Or, using our first result

t = x / v{x},

makes the equation into:

y{0} = g x² / (2 v{0}²).

Hope that makes sense,
~W.O.M.B.A.T.

Answer 3

First you must remember that projectile motion, which this is, has two components. There is an x-component , or horizontal component, which neglecting air resistance, is constant. And there is a y-component which obeys the Law of Falling Bodies.
So this snowball has a horizontal speed that starts of at 11.4 m/s and this stays the same until it hits the ground 33 meters from the bottom of the building. Ask yourself, "How many seconds did it take to travel 33 meters horizontally traveling 11.4 m/s in the horizontal direction?'
The solution is simply t=x/v = (33m)/(11.4m/s)= 2.89 s

Now in order to find the height of the building use the falling body equation: y=(1/2)gt^2= (4.9)(2.89)^2=41.1 m

Remember that the two motions are totally independent of one another!

Answer 4

This is a rate problem in a couple of dimensions....

First, assume that v will be constant. That makes things a lot simpler.

Next find out how much time it takes for an object traveling 11.4 m/s to go 33 m.

Finally, use your acceleration formulas to find out how far something falls in the time you calculated above.

Fire Pythagoras. You don't need him for this problem.