Answer:
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.
