A ball rolls horizontally off a 3.2 m high table and lands 2.2 m from the end of the table. What is the horizontal velocity of the ball when it leaves the table?
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To find the horizontal velocity of the ball when it leaves the table, we can use the principles of projectile motion.
The key idea here is that horizontal and vertical motion are independent of each other. This means that the horizontal velocity remains constant throughout the motion, while the vertical velocity changes due to the effect of gravity.
First, let's analyze the vertical motion of the ball. We know that the ball is dropped from a height of 3.2 m and lands 2.2 m away from the edge of the table. Therefore, the total horizontal distance traveled by the ball is 2.2 m.
Using the equation for vertical displacement, we can relate the initial vertical velocity, time of flight, and acceleration due to gravity:
Δy = v0y * t + (1/2) * g * t^2
Since the ball is dropped, the initial vertical velocity is 0, and the equation simplifies to:
Δy = (1/2) * g * t^2
Plugging in the values, we have:
3.2 m = (1/2) * 9.8 m/s^2 * t^2
Simplifying the equation, we get:
9.8 t^2 = 6.4
Dividing both sides by 9.8:
t^2 = 0.653
Taking the square root of both sides:
t = 0.808 s
Now that we have the time of flight, we can find the horizontal velocity using the horizontal distance traveled:
d = v0x * t
Since the horizontal velocity remains constant throughout the motion, v0x is the horizontal velocity we want to find. Rearranging the equation, we have:
v0x = d / t
Plugging in the values, we get:
v0x = 2.2 m / 0.808 s
Calculating the value:
v0x ≈ 2.72 m/s
Therefore, the horizontal velocity of the ball when it leaves the table is approximately 2.72 m/s.