A small steel ball rolls off a table (1 meter high) with a speed of 2 m/s. Where should they place a can (15 cm high; diameter 10 cm) to catch the ball?

Xo = 2 m/s.

h = 0.5g*t^2 = 1 m.
4.9t^2 = 1
t^2 = 0.204
Tf = 0.452 s. = Fall time.

Dh = Xo * Tf = 2m/s * 0.452 = 0.904 m =
Hor. distance from table.

d = Dh - r = 0.904 - 0.05 = 0.854 m
= Hor. distance from table.

To determine where to place the can to catch the rolling ball, we need to calculate the horizontal distance traveled by the ball before it hits the ground. Here's how you can do it:

Step 1: Calculate the time it takes for the ball to reach the ground.
We can use the kinematic equation for vertical motion:
h = (1/2) * g * t^2 + v_i * t + h_i
Where:
h = vertical distance (in this case, 1 meter)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken
v_i = initial vertical velocity (in this case, 0 m/s since the ball rolls horizontally)
h_i = initial vertical height (in this case, 1 meter)

Rearranging the equation, we get:
0 = -4.9t^2 + 0t + 1
Simplifying, we have:
4.9t^2 = 1
t^2 = 1 / 4.9
t ≈ 0.45 seconds

Step 2: Calculate the horizontal distance traveled by the ball during that time.
Using the equation:
d = v * t
Where:
d = horizontal distance
v = horizontal velocity (in this case, the velocity of the ball rolling off the table - 2 m/s)
t = time taken (0.45 seconds)

Substituting the values, we get:
d = 2 * 0.45
d ≈ 0.9 meters

Therefore, the ball will travel approximately 0.9 meters horizontally before hitting the ground. To place the can in a position to catch the ball, it should be placed 0.9 meters away from the edge of the table in the horizontal direction.