Two spheres are launched horizontally from a 1.2 m high table. Sphere A is launc

Question

Two spheres are launched horizontally from a 1.2 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 2.5 m/s.
Part A What is the time for the sphere A to hit the floor?
Part b What is the time for the sphere B to hit the floor?
Part c What is the distance that sphere a travels from the edge of the table?
Part d What is the distance that sphere B travels from the edge of the table?
Two spheres are launched horizontally from a 1.2 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 2.5 m/s.
Part A What is the time for the sphere A to hit the floor?
Part b What is the time for the sphere B to hit the floor?
Part c What is the distance that sphere a travels from the edge of the table?
Part d What is the distance that sphere B travels from the edge of the table?

Part A What is the time for the sphere A to hit the floor?
Part b What is the time for the sphere B to hit the floor?
Part c What is the distance that sphere a travels from the edge of the table?
Part d What is the distance that sphere B travels from the edge of the table?

Explanation / Answer

To answer this question, the first thing to remember is that acceleration problems can be seperated into horizontal and vertical components.

The next thing to do is to establish a coordinate system. Normally, we consider up as positive so up is positive. The y zero coordinate will be the floor and the x zero coordinate will be the edge of the table. Positive x will the direction the ball is going when it leaves the table.

Notice that the first equation involves time so we need an acceleration equation that includes time. This equation would be:

y = y(0) + V(0)*t + 0.5*g*t^2; g = -9.81 m/s^2 = gravity which is the vertical force acting on the ball when it leaves the table. y = 0, y(0) = 1.2m, and V(0) = 0 since the table is horizontal and we are looking for the time to fall a vertical distance. Rearrange the equation to solve for the unknown which is time:
y = y(0) + V(0)*t + 0.5*g*t^2; V(0) = 0
[y - y(0)] / {0.5*g) = t^2
t^2 = [y - y(0)] / {0.5*g)
t = sqrt{ [y - y(0)] / {0.5*g)}
Plugging in the known numbers gives you:
t = sqrt[(0-1.2m) / (0.5 * -9.81m/s^2)]
t = 0.495 seconds = time for sphere A to hit the floor (Part A)

Since Sphere B starts from the same height with the same vertical velocity as sphere A, the time to fall the vertical distance will be the same for sphere B as it is for sphere A. Therefore t = 0.495 seconds for Sphere B (Part B)

Use the same equation to calculate horizontal distances; however, this time the unknown you need to solve for distance [X-X(0)] and not t. For the value of t, you will use the t found in parts A and B of this problem. Starting with the equation:
x = x(0) + V(0)*t + 0.5*g*t^2; V(0) = 4.5 m/s, g = - 9.81m/s^2, and t = 0.495s. Rearrange the equation to solve for [X-X(0)]:
X - X(0) = V(0)*t + 0.5*g*t^2
Plugging in the numbers gives:
X - X(0) = 4.5m/s * 0.495s + 0.5 * -9.81 * (0.495s)^2
X - X(0) = 2.2275m - 1.2018m = 1.0257 m = distance Sphere A travels from the edge of the table (Part C)

Repeat the calculations done in Part C for Part D. This gives:
X - X(0) = V(0)*t + 0.5*g*t^2
X - X(0) = 2.5m/s * 0.495s + 0.5 * -9.81 * (0.495s)^2
X - X(0) = 1.2375m - 1.2018m = 0.0357 m = distance Sphere B travels from the edge of the table (Part D)

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