Aball is traveling uphill with an initial velocity of 5.0 m/s and an acceleration of -2.0 m/s^2. a) how fast is the ball traveling when it returns to its release point?

b) where is the ball's maximum uphill position? this is the point where v0 is 0m/s

c) the ball reaches a point 6.0 meters uphill from the release point twice once on the way up and once on the way down what is the ball's velocity at each of those moments remember that on the way up v is 5m/s and on the way down v is 0m/s

A) The ball is traveling at 5.0 m/s (magnitude) when the ball returns to its release point.

B) The maximum uphill position is at 6.25 m from the release point.

C) On the way up, the velocity of the ball at x = 6.0 m is 1 m/s and on the way down it is - 1m/s.

Explanation:

Hi there!

The position and velocity of the ball can be calculated using the following equations:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of th ball at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time.

v = velocity at time t.

A) Let´s place the origin of the frame of reference at the point at which the ball has a velocity of 5.0 m/s. Then, x0 = 0.

When the ball returns to the initial point, its position will be 0. Then using the equation of position we can calculate at which time the ball is at x = 0:

x = x0 + v0 · t + 1/2 · a · t²

0 m = 5.0 m/s · t - 1/2 · 2.0 m/s² · t²

0 m  = 5.0 m/s · t - 1.0 m/s² · t²

0 m = t (5.0 m/s - 1.0 m/s² · t)

t = 0 (this is logic becuase the ball starts at x = 0)

and

5.0 m/s - 1.0 m/s² · t = 0

t = -5.0 m/s / -1.0 m/s²

t = 5.0 s

With this time, we can calculate the velocity of the ball:

v = v0 + a · t

v = 5.0 m/s - 2.0 m/s² · 5.0 s

v = -5.0 m/s

The ball is traveling at 5.0 m/s when the ball returns to its release point.

B) Let´s use the equation of velocity to obtain the time at which the ball is at its maximum uphill position:

v = v0 + a · t

0 = 5.0 m/s - 2.0 m/s² · t

-5.0 m/s/ -2.0 m/s² = t

t = 2.5 s

Now, using the equation of position, let´s find the position of the ball at t = 2.5 s. This position will be the maximum uphill position because at that time the velocity is 0:

x = x0 + v0 · t + 1/2 · a · t²

x = 5.0 m/s · 2.5 s - 1/2 · 2.0 m/s² · (2.5 s)²

x = 6.25 m

The maximum uphill position is at 6.25 m from the release point.

C) First, let´s find the time at which the ball is 6.0 meters uphill from the releasing point:

x = x0 + v0 · t + 1/2 · a · t²

6.0 m = 5.0 m/s · t - 1/2 · 2 m/s² · t²

0 = -1 m/s² · t² + 5.0 m/s · t - 6.0 m

Solving the quadratic equation using the quadratic formula:

a = -1

b = 5

c = -6

t = [-b ± √(b² - 4ac)]/2a

t₁ = 2 s (on its way up)

t₂ = 3 s (on its way down)

Now, let´s calculate the velocity of the ball at those times:

v = v0 + a · t

v = 5.0 m/s - 2 m/s² · 2 s = 1 m/s

v = 5.0 m/s - 2 m/s² · 3 s = -1 m/s

On the way up, the velocity of the ball at x = 6.0 m is 1 m/s and on the way down it is - 1m/s.

which statement about this sum is true? −245+2.7

as technology advances, new experiments often expose problems in accepted theories. c

hope i

-lavira: )