An orange rolls across a table with a speed of 0.5 m/s and falls off the edge. If it lands 3 m from the edge of the table, from what height does it fall?

d = Vx*Tf = 3 m.

0.5m/s * Tf = 3
Tf = 6 s. = Fall time.

h = 0.5g*Tf^2 = 4.9 * 6^2 = 176.4 m.

Given:

Vox= 0.5m/s
Dx= 3m
g= 9.8 m/s²

solve for
dy

Formula: dy=voy+ 1/2 gt²
solve for time
dx=vxt
dx/vx=vxt/vx
(cancel both vx)
t=dx/vx
t= 3m/0.5m/s

formula dy=voy + t 1/2gt²

dy=1/2(9.8 m/s²) (6s)²

answer: -176.4 m

A 25-kg boy is riding a 12-kg bike moving at4.5m/s.what is the total momentum of the boy and the bike together

Well, let me juggle with this question for a moment. If the orange lands 3 meters from the edge of the table, it means it took some time to reach there. Now, gravity might not find it funny, but it always pulls objects down at an acceleration of about 9.8 m/s². So, using some impressive calculations, we can determine that the orange was in the air for approximately 0.306 seconds.

Now, during this time, the orange was rolling at a speed of 0.5 m/s, which is about as fast as a tortoise on roller skates. Since speed is equal to distance divided by time, we can determine that the orange traveled a distance of 0.153 meters during its airborne journey.

If we visualize this as a triangle, with the height being the distance the orange fell, the base being the distance it traveled horizontally, and the time as the hypotenuse, we can apply some acrobatics to find that the height is approximately 0.097 meters.

So, the orange fell from a height of about 0.097 meters. Now, if only it had a chance to try out for the circus with that incredible stunt!

To determine the height from which the orange falls, we can use the equations of motion along with the principle of conservation of energy.

First, let's consider the horizontal motion of the orange. Since its initial velocity is 0.5 m/s and it lands 3 m from the edge, we can use the equation of motion for uniform motion:

distance = speed × time

In this case, the distance is 3 m and the speed is 0.5 m/s. Rearranging the formula, we can solve for the time it takes the orange to reach the edge of the table:

time = distance / speed
= 3 m / 0.5 m/s
= 6 s

Now, let's consider the vertical motion of the orange when it falls. We can use the equation for the height of an object in free fall from rest:

height = (1/2) × gravity × time^2

Here, the time is 6 s and we need to determine the value of gravity. Gravity on Earth is approximately 9.8 m/s^2. Applying these values to the equation, we can solve for the height from which the orange falls:

height = (1/2) × 9.8 m/s^2 × (6 s)^2
= 1/2 × 9.8 m/s^2 × 36 s^2
= 1/2 × 9.8 m/s^2 × 1296 s^2
= 4.9 m/s^2 × 1296 s^2
= 6350.4 m

Therefore, the orange falls from a height of approximately 6350.4 meters.

An orange rolls across a table with a speed of 0.5 m/s and falls off the edge. If it lands 3 m from the edge of the table, from what height does it fall?