tennis ball rolls off the edge of a tabletop 0.900m above the floor and strikes the floor at a point 1.60m horizontally from the edge of the table.

A: Find the time of flight of the ball

B: Find the magnitude of the initial velocity of the ball.

C: Find the magnitude of the velocity of the ball just before it strikes the floor.

D: Find the direction of the velocity of the ball just before it strikes the floor.

A. h = 0.5g*t^2 = 0.90 m.

4.9t^2 = 0.90
t^2 = 0.184
t = 0.429 s. = Fall time.

B. Dx = Xo*t = 1.6 m.
Xo * 0.429 = 1.6
Xo = 3.73 m/s. = Hor. component.

C. Y = Yo+g*t = 0 + 9.8*0.429 = 4.20 m/s
= Ver. component.

V^2 = Xo^2 + Y^2 = 3.73^2 + 4.2^2=31.55
V = 5.62 m/s.

D. tan A = Y/Xo = 4.2/3.73 = 1.12601
A = 48.4o. = Direction.

Distance covered along the pathway,

s = sqrt(0.9^2 + 1.6^2) = 1.836 m
Now, s = ut + ½*gt^2
=> 1.836 = 0.5 * 9.8 * t^2 [since u=0]
=> t = 0.612 sec (Fall time)

A: Well, the time of flight of the ball can be determined using projectile motion equations. First, we can find the time it takes for the ball to fall vertically to the floor using the equation h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity, and t is the time. Rearranging the equation, we have t = sqrt(2h/g). Plugging in the values, we get t = sqrt(2 * 0.9 / 9.8) = 0.427 seconds.

B: To find the magnitude of the initial velocity, we can use the horizontal distance and time of flight. The horizontal distance can be found using the equation d = v_i * t, where d is the distance, v_i is the initial velocity, and t is the time. Rearranging the equation, we have v_i = d / t. Plugging in the values, we get v_i = 1.6 / 0.427 = 3.748 m/s.

C: The magnitude of the velocity of the ball just before it strikes the floor can be found using the equation v_f = v_i + at, where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and t is the time. Since there is no horizontal acceleration, the vertical acceleration due to gravity is the only acceleration acting on the ball. Plugging in the values, we get v_f = 0 + (9.8 * 0.427) = 4.1866 m/s.

D: The direction of the velocity of the ball just before it strikes the floor is straight down or vertically downwards. So, if you ever find yourself underneath the ball, make sure to watch your head and dodge it like a pro!

To solve this problem, we can use the equations of motion for a freely falling object. Let's assume the positive direction is upwards.

A: Find the time of flight of the ball:

The vertical distance traveled by the ball can be found using the equation: h = ut + (1/2)gt^2, where h is the vertical distance (0.900m), u is the initial vertical velocity (unknown), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time of flight.

Plugging in the known values, we have: 0.900 = ut + (1/2)(-9.8)t^2.

Simplifying the equation, we get: 0.900 = (1/2)(-9.8)t^2.

Rearranging the equation, we have: -4.9t^2 = 0.900.

Dividing both sides by -4.9, we get: t^2 = -0.900 / -4.9.

Taking the square root of both sides, we get: t = √(0.900 / 4.9).

Calculating the value, we find: t ≈ 0.433 seconds.

Therefore, the time of flight of the ball is approximately 0.433 seconds.

B: Find the magnitude of the initial velocity of the ball:

Using the equation: v = u + gt, where v is the final velocity (0 when the ball hits the floor), u is the initial vertical velocity (unknown), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time of flight (0.433s).

Plugging in the known values, we have: 0 = u + (-9.8)(0.433).

Simplifying the equation, we get: 0 = u - 4.244.

Rearranging the equation, we find: u = 4.244 m/s.

Therefore, the magnitude of the initial velocity of the ball is approximately 4.244 m/s.

C: Find the magnitude of the velocity of the ball just before it strikes the floor:

Since there is no horizontal force acting on the ball, the horizontal component of its velocity remains constant. Therefore, the horizontal component of the final velocity is equal to the horizontal component of the initial velocity.

Using the equation: v = u + gt, where v is the final velocity (unknown), u is the initial vertical velocity (4.244 m/s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time of flight (0.433s).

Plugging in the known values, we have: v = 4.244 + (-9.8)(0.433).

Calculating the value, we find: v ≈ -0.918 m/s.

Therefore, the magnitude of the velocity of the ball just before it strikes the floor is approximately 0.918 m/s.

D: Find the direction of the velocity of the ball just before it strikes the floor:

The negative sign in the magnitude of the velocity indicates that the ball is moving downwards just before it strikes the floor. Therefore, the direction of the velocity of the ball just before it strikes the floor is downwards.

To solve these questions, we can use the principles of projectile motion. Here's how you can find the answers to each part:

A: Find the time of flight of the ball
The time of flight is the total time it takes for the ball to reach the floor. We can use the equation of motion for vertical displacement:

y = viy * t - (1/2) * g * t^2

Where:
- y is the vertical displacement (0.900 m)
- viy is the vertical component of the initial velocity (unknown)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time of flight (unknown)

Since the ball starts at rest vertically, the initial vertical velocity (viy) is 0 m/s. Therefore, the equation simplifies to:

y = - (1/2) * g * t^2

Plug in the known values and solve for t.

B: Find the magnitude of the initial velocity of the ball
To find the magnitude of the initial velocity, we need to find the horizontal component (vix) and vertical component (viy) separately. Since the horizontal velocity is constant throughout the motion, we can use the equation:

x = vix * t

Where:
- x is the horizontal displacement (1.60 m)
- vix is the horizontal component of the initial velocity (unknown)
- t is the time of flight (found in part A)

Plug in the known values and solve for vix.

The magnitude of the initial velocity can be found using the Pythagorean theorem:

|v| = sqrt(vix^2 + viy^2)

Substitute the values of vix and viy into the equation and solve for |v|.

C: Find the magnitude of the velocity of the ball just before it strikes the floor
The horizontal component of velocity remains constant throughout the motion, so it is the same as the initial horizontal component of velocity (vix). The vertical component of velocity (viy) changes due to the acceleration due to gravity (g). To solve for the magnitude of the velocity just before it strikes the floor, we can use the equation from part B:

|v| = sqrt(vix^2 + viy^2)

Substitute the known values of vix and viy (found in part A) into the equation and solve for |v|.

D: Find the direction of the velocity of the ball just before it strikes the floor
The direction of velocity can be determined by analyzing the components of velocity (vix and viy) separately. Since the horizontal component of velocity remains constant throughout the motion, the direction is horizontal, which is typically considered to be along the x-axis. The vertical component of velocity (viy) changes due to gravity, so its direction is pointing downwards, usually considered negative on the y-axis. Therefore, the direction of the velocity just before it strikes the floor is the vector sum of the horizontal and vertical components.

I hope this explanation helps you solve the problem!