Ball X is dropped from a height of 1.25 meters above the ground. At the same time and from exactly the same height, Ball Y is shot out horizontally at an initial velocity of 8.0 m/s. Air resistance is negligible and assume g =-10 m/s?
How far away from Ball X will Ball Y hit the ground?

The distance between Ball X and Ball Y is 4.0 meters

Explanation:

First we have to use the information we have from Ball Y to find time as well as it’s distance.

Ball Y’s Given:

Vx=8.0m/s

ax=0m/s^2

x=?

y=1.25m

ay=-10m/s^2

To find the time, use the equation t=(2y/g)^1/2.

t=(2(1.25)/10)^1/2

t=(2.5/10)^1/2

t=(0.25)^1/2

t=0.5s

Now we have to calculate how far away Ball Y fell from the base of the cliff. To do this use the equation x=(Vx)(t) and plug in the values.

x=(8.0m/s)(0.5s)

x=4.0m

Because Ball X is dropped from the cliff, we can conclude that the ball is 0meters away from the cliff. With this in mind, we can assume that the distance between Ball X and Ball Y is 4.0m.

Total Distance = Ball Y - Ball X

Total Distance = 4.0 m - 0 m

Total Distance = 4.0 m

need to see circuit to get time constant

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