A tennis ball rolls off the edge of a tabletop 0.800m above the floor and strikes the floor at a point 1.10m horizontally from the edge of the table.

1) Find the time of flight of the ball.
2) Find the magnitude of the initial velocity of the ball.
3) Find the magnitude of the velocity of the ball just before it strikes the floor.
4) Find the direction of the velocity of the ball just before it strikes the floor.

t=.404, Vi=2.75 m/s, Vf=3.92m/s

To solve these problems, we can use the equations of motion for a projectile. Here are the steps to find the answers to each question:

1) Find the time of flight of the ball:
To find the time of flight, we can use the equation for vertical motion. The ball starts from rest, and the only force acting on it is gravity.

The equation we can use is:
y = v0*t - (1/2)gt^2

where:
y is the vertical displacement (0.800m)
v0 is the initial vertical velocity (unknown)
t is the time of flight (unknown)
g is the acceleration due to gravity (9.8 m/s^2)

Since the ball starts from rest, the initial vertical velocity (v0) is 0. Therefore, the equation simplifies to:
0.800 = - (1/2)gt^2

Now, we can solve for t. Rearranging the equation:
t^2 = (2 * 0.800) / g

t^2 = 0.1633

Taking the square root of both sides:
t = 0.404 s (rounded to 3 decimal places)

So, the time of flight of the ball is 0.404 seconds.

2) Find the magnitude of the initial velocity of the ball:
To find the initial velocity, we can use the horizontal motion of the ball. The ball covers a horizontal distance of 1.10m.

The equation we can use is:
x = v0x * t

where:
x is the horizontal distance (1.10m)
v0x is the initial horizontal velocity (unknown)
t is the time of flight (0.404s)

Rearranging the equation:
v0x = x / t

v0x = 1.10 / 0.404

v0x = 2.723 m/s (rounded to 3 decimal places)

So, the magnitude of the initial velocity of the ball is 2.723 m/s.

3) Find the magnitude of the velocity of the ball just before it strikes the floor:
The vertical velocity of the ball just before it strikes the floor can be found using the equation for vertical motion.

The equation we can use is:
v = v0 - gt

where:
v is the final vertical velocity (unknown)
v0 is the initial vertical velocity (0 m/s)
g is the acceleration due to gravity (9.8 m/s^2)
t is the time of flight (0.404s)

Rearranging the equation:
v = -gt

Substituting the values:
v = -(9.8)(0.404)

v = -3.9632 m/s (rounded to 4 decimal places)

The magnitude of the velocity is always positive, so:
|v| = 3.9632 m/s (rounded to 4 decimal places)

So, the magnitude of the velocity of the ball just before it strikes the floor is 3.9632 m/s.

4) Find the direction of the velocity of the ball just before it strikes the floor:
Since the vertical velocity is negative, it means the ball is moving downward. The horizontal velocity, v0x, remains constant throughout the motion. Therefore, the direction of the velocity just before it strikes the floor is the same as the initial horizontal velocity, which is to the right.

So, the direction of the velocity of the ball just before it strikes the floor is to the right or horizontally.

I will be happy to critique your thinking.

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