A 75 g autographed baseball slides off of a

1.3 m high table and strikes the floor a hori-
zontal distance of 0.65 m away from the table.
The acceleration of gravity is 9.81 m/s2 .

What was the direction of the ball’s velocity
just before it hit the floor?
That is, at what angle (in the range −90◦ to
+90◦ relative to the horizontal directed away
from the table) did the ball hit the floor?
Answer in units of ◦.

To find the direction of the ball's velocity just before it hit the floor, we can use the principles of projectile motion. The velocity of the ball just before hitting the floor can be broken down into horizontal and vertical components.

Given:
Mass of the ball (m) = 75 g = 0.075 kg
Height of the table (h) = 1.3 m
Horizontal distance (x) = 0.65 m
Acceleration due to gravity (g) = 9.81 m/s^2

First, let's calculate the time taken for the ball to fall from the table to the floor:
Using the equation h = 0.5 * g * t^2, where h is the height, g is the acceleration due to gravity, and t is time, we can rearrange the equation to solve for time (t):
1.3 = 0.5 * 9.81 * t^2
Simplifying, t^2 = (1.3 * 2) / 9.81
t^2 = 0.2649
Taking the square root of both sides, t ≈ 0.5157 s

Next, we can calculate the horizontal component of the velocity (Vx):
Using the equation x = Vx * t, where x is the horizontal distance and t is the time,
0.65 = Vx * 0.5157
Vx = 0.65 / 0.5157
Vx ≈ 1.2604 m/s

Now, we can calculate the vertical component of the velocity (Vy) just before hitting the floor:
Using the equation Vy = g * t, where g is the acceleration due to gravity and t is the time,
Vy = 9.81 * 0.5157
Vy ≈ 5.0606 m/s

Finally, we can find the angle (θ) of the ball's velocity just before it hit the floor:
Using the equation tan(θ) = Vy / Vx,
tan(θ) = (5.0606) / (1.2604)
θ = atan(5.0606 / 1.2604)
θ ≈ 75.76 degrees

Therefore, the ball's velocity just before it hit the floor has an angle of approximately 75.76 degrees relative to the horizontal directed away from the table.