An Iraqi reporter throws his shoe at President Bush during the Presiden'ts last press conference in Bagdad.

The reporter is a horizontal distance of 12 feet from the President, and launches his shoe from a height of 3 feet. At lift-off, the shoe has a speed of 24 feet per second, and its trajectory makes an angle of 45 with the horizontal. The only force acting on the shoe during its flight is gravity, which produces a downward acceleration of 32 feet/ sec^2.

How high will the shoe be above the ground when it is over the position where th epresiden is standing?

The shoe will be 6.251 ft above the ground when it reaches the president.

Step-by-step explanation:

Let's first find the horizontal and vertical velocity of the shoe.

Horizontal velocity: 24 * Cos(45°) = 16.97 ft/s

Vertical velocity: 24 * Sin(45°) = 16.97 ft/s

Travel time to cover distance between president and reporter:

Travel time = Distance / horizontal velocity

Travel time = 12 / 16.97

Travel time = 0.707 seconds

Now that we know the time we can solve the following motion equation to solve for change in vertical distance after the shoe is thrown:

here the initial speed (u) is 16.97 ft/s

the acceleration (a) is -32ft/s^2

and the time is 0.707 seconds

Solving for s we get:

s = 11.998 - 8.747

s = 3.251 ft

Since the shoe was already 3 ft above the ground when we started, we add that to the change in distance:

3 + 3.251 = 6.251 ft

So the shoe will be 6.251 ft above the ground when it reaches the president.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.