A ball rolls with a speed of 6.4 m/s toward the edge of a table that is 0.7 m above the floor. The ball rolls off the table.
a. Find the magnitude of the balls velocity when it hits the floor. m/s
b. At what angle, measured from the horizontal, will the ball's velocity make with the ground? °
The ball after leaving the table
takes part in two motions: horizontal uniform motion
L = v(hor) •t,
and vertical free fall
H = gt^2/2,
t = sqroot(2H/g) = sqroot(2•0.7/9.8) = 1.43 s.
v(vert) = gt = 9.8•1.43 = 14 m/s.
tan α = v(vert)/ v(hor)=14/6.4=2.19
α = 65.4 degr
SORRY!
t = sqroot(2H/g) = sqroot(2•0.7/9.8) = 0.143 s.
v(vert) = gt = 9.8•0.143 = 1.4 m/s.
tan α = v(vert)/ v(hor)=1.4/6.4=0.219
α = 12.35 degr
To solve this problem, we need to analyze the motion of the ball using the principles of physics. We'll consider two main components of the ball's motion: horizontal and vertical.
a. To find the magnitude of the ball's velocity when it hits the floor, we need to determine the horizontal and vertical components of its velocity separately.
1. Horizontal component: The ball rolls with a constant speed, so its horizontal velocity remains constant. We can find it by using the formula: horizontal velocity = initial horizontal velocity = final horizontal velocity.
The given speed of the rolling ball is 6.4 m/s. Since there is no acceleration in the horizontal direction, the horizontal component of the ball's velocity remains 6.4 m/s.
2. Vertical component: The ball experiences vertical free fall acceleration due to gravity (9.8 m/s²), as it falls from the table.
To find the vertical component of the ball's velocity, we'll use the equation of motion in the vertical direction:
final vertical velocity ² = initial vertical velocity ² + 2 * acceleration * displacement
The initial vertical velocity is 0 since the ball starts falling from rest, and the displacement is the height of the table (0.7 m).
final vertical velocity ² = 0 + 2 * (-9.8 m/s²) * (-0.7 m)
final vertical velocity ² = 13.72 m²/s²
final vertical velocity = √(13.72 m²/s²) ≈ 3.71 m/s (rounded to 2 decimal places)
Now, to find the magnitude of the ball's velocity when it hits the floor, we'll combine the horizontal and vertical components using the Pythagorean theorem:
velocity = √(horizontal velocity ² + vertical velocity ²)
velocity = √((6.4 m/s) ² + (3.71 m/s) ²)
velocity ≈ √(41.28 m²/s² + 13.72 m²/s²)
velocity ≈ √55 m²/s²
velocity ≈ 7.42 m/s (rounded to 2 decimal places)
Therefore, the magnitude of the ball's velocity when it hits the floor is approximately 7.42 m/s.
b. Now, let's find the angle at which the ball's velocity makes with the ground.
angle = arctan(vertical velocity / horizontal velocity)
angle = arctan(3.71 m/s / 6.4 m/s)
angle ≈ arctan(0.58)
angle ≈ 29.5° (rounded to 1 decimal place)
Therefore, the angle at which the ball's velocity makes with the ground is approximately 29.5°.