Two Spheres Launched Horizontally

[Raimuna]

Homework Statement

Two spheres are launched horizontally from a 1.1 m-high table. Sphere A is launched with an initial speed of 4.5 m/s. Sphere B is launched with an initial speed of 1.5 m/s.
-What is the time for the sphere A to hit the floor?
-What is the distance that sphere A travels from the edge of the table?

Homework Equations

Speed=total distance/total time

The Attempt at a Solution

I have tried the speed formula, but I did incorrectly.
total time = 1.1/4.5
= 0.24 s

 

[evanlee]

The trick here is that it is being launched horizontally -- meaning all of the initial velocity is in the x direction. Since the object starts at a height of 1.1m, it must drop that distance.

Notice that the only force acting on the sphere is gravity. When you look at the sphere's motion in the y direction, it will only be effected by this force, not the initial velocity. This means that, in terms of the y component of the motion, the sphere is in free fall.

Using the fact that it starts at 1.1m, you can come up with the time it takes to fall.

remember: h = 1/2 g*t^2

solving for t: t= (2h/g)^(1/2)

Now that you have the time that the sphere is in motion, it is not too difficult to come up with the distance that it travels. As I said before, the initial velocity is completely in the x direction. We can assume that this velocity remains constant for the entirety of the motion, and thus the distance traveled is equal to the velocity multiplied by the time traveled.

t was calculated before, so d=vt, where v = 4.5 m/s

hope this helps!

 

[evanlee]

adding to my last post, if you are being asked to find the total displacement from the original launching point of the sphere, you will need to use the pythagorean theorem.

We know that the sphere has fallen 1.1m in the y direction. By solving for d as described above (t = 0.47 s) we get d = 2.13m.

Now by the Pythagorean Theorem:

r = (x^2 + y^2)^(1/2)

this approximately comes out to 2.4m