C Exercise 3.57

In laboratory situations, a projectile’s range can be used to determine its speed. To see how this is done, suppose a ball rolls off a horizontal table and lands 1m out from the edge of the table.
Part A
If the tabletop is 70cm above the floor, determine the time the ball is in the air.
Express your answer using two significant figures.
t =
0.38
s

Part B
Determine the ball's speed as it left the table top.
Express your answer using two significant figures.

v =
1.84
m/s
cant figure out part b

A. h = 0.5g*t^2 = 0.7 m.

4.9t^2 = 0.7
t^2 = 0.1429
Tf = 0.378 s. = Fall time.

B. Dx = Xo*Tf = 1 m
Xo * 0.378 = 1
Xo = 1/0.378 = 2.65 m/s. = Initial hor.
velocity.

To determine the ball's speed as it left the tabletop, you can use the range formula for projectile motion. The formula for the horizontal range (R) of a projectile is given by:

R = (initial velocity * time of flight)

In this case, the range of the projectile is given as 1m, and the time of flight (t) is determined in Part A as 0.38s. Rearranging the formula, you can solve for the initial velocity (v):

v = R / t

Using the given values:

v = 1m / 0.38s ≈ 2.63 m/s

Now, since the answer should be expressed using two significant figures, the final answer for the ball's speed as it left the tabletop is approximately:

v = 2.63 m/s (rounded to two significant figures)

To determine the ball's speed as it left the table top in Part B, we can use the formula for range:

Range = (initial velocity * time) + (0.5 * acceleration * time^2)

Since the ball rolls off a horizontal table, we can assume that it had an initial vertical velocity of 0 (since it falls straight down due to gravity). Therefore, the vertical displacement is given by:

Vertical displacement = 70 cm = 0.7 m

The initial vertical velocity is 0 m/s, and the acceleration due to gravity is approximately 9.8 m/s^2.

The time the ball is in the air can be calculated from the formula for vertical displacement:

Vertical displacement = (initial vertical velocity * time) + (0.5 * acceleration * time^2)
0.7 = (0 * t) + (0.5 * 9.8 * t^2)

Simplifying the equation gives:

0.7 = 4.9 * t^2

Now, we can solve for t by rearranging the equation:

t^2 = 0.7 / 4.9
t^2 = 0.14285
t ≈ √(0.14285)
t ≈ 0.37857

Since we need to express the answer using two significant figures, we round the time to:

t ≈ 0.38 s

Therefore, the time the ball is in the air is approximately 0.38 seconds.

To determine the ball's speed as it left the table top, we can use the formula for horizontal distance:

Horizontal distance = initial horizontal velocity * time

The horizontal distance is given as 1 m, and we've calculated the time to be approximately 0.38 s.

1 = initial horizontal velocity * 0.38

Solving for the initial horizontal velocity gives:

initial horizontal velocity = 1 / 0.38

Calculating this value gives:

initial horizontal velocity ≈ 2.63 m/s

Since we need to express the answer using two significant figures, we round the initial horizontal velocity to:

initial horizontal velocity ≈ 2.6 m/s

Therefore, the ball's speed as it left the table top is approximately 2.6 m/s.