Bob throws a ball straight up with an initial speed of 50 feet per second from a height of 6 feet.

(a) Find parametric equations that describe the motion of the ball as a function of time. (Do this on paper. Your instructor may ask you to turn in this work.)
(b) How long is the ball in the air?
seconds
(c) When is the ball at its maximum height?
seconds
Determine the maximum height of the ball.
feet
(d) Simulate the motion of the ball by graphing the equations found in part (a).

(a) To find the parametric equations that describe the motion of the ball, we need to consider the equations for position in terms of time, velocity, and acceleration.

Let's assume the positive direction is upwards. We can represent the initial position of the ball as (0,6) since it starts from a height of 6 feet.

The velocity of the ball, v(t), is given by:

v(t) = v0 - g * t

where:
v0 = initial velocity = 50 ft/s
g = acceleration due to gravity = -32 ft/s^2 (negative sign since it's acting in the opposite direction)

Integrating v(t), we get:

s(t) = ∫(v(t)) dt
= ∫(v0 - g * t) dt
= v0 * t - (1/2) * g * t^2

This gives us the position s(t) of the ball as a function of time.

The parametric equations for the motion of the ball are:
x(t) = 0
y(t) = v0 * t - (1/2) * g * t^2 + 6

(b) The ball is in the air until it hits the ground, which occurs when y(t) = 0. So we can solve the equation y(t) = 0 to find the time when the ball hits the ground.

v0 * t - (1/2) * g * t^2 + 6 = 0

Simplifying this equation, we have:

-16t^2 + 50t + 6 = 0

Using the quadratic formula, we solve for t:

t = (-b ± sqrt(b^2 - 4ac))/(2a)
= (-(50) ± sqrt((50)^2 - 4(-16)(6)))/(2(-16))

Calculating, we find two possible solutions: t ≈ 3.03s and t ≈ 0.14s. However, the time when the ball hits the ground should be a positive value, so t ≈ 3.03s is the correct solution.

Therefore, the ball is in the air for approximately 3.03 seconds.

(c) To find the time when the ball is at its maximum height, we need to find the vertex of the parabolic path. The vertex occurs at the maximum point of the curve.

The time of the maximum height can be found using the equation t = -b / (2a).

Using this formula, we have:

t = -(50) / (2(-16))
≈ 1.563s

Therefore, the ball is at its maximum height around 1.563 seconds.

(d) To simulate the motion of the ball by graphing the equations, we plot the parametric equations:

x(t) = 0
y(t) = v0 * t - (1/2) * g * t^2 + 6

On a graph, we can plot the time (t) on the x-axis and the position (y) on the y-axis. The x-coordinate is constant at 0 since the ball is thrown straight up. And the y-coordinate is given by the equation.

By plotting this graph, we can visualize the motion of the ball.

To find the parametric equations that describe the motion of the ball in terms of time, we can use the following kinematic equations:

1. Vertical position equation:
y = h + v₀t - 16t²

2. Vertical velocity equation:
y' = v₀ - 32t

For this problem, we are given the initial speed, v₀, as 50 feet per second, and the initial height, h, as 6 feet. Let's proceed with solving part (a) first:

(a) Find parametric equations that describe the motion of the ball as a function of time:

1. Vertical position equation:
Since we know the initial position is 6 feet (h = 6), substitute these values into the vertical position equation:
y = 6 + 50t - 16t²

2. Vertical velocity equation:
Substitute the initial velocity of 50 feet per second (v₀ = 50) into the vertical velocity equation:
y' = 50 - 32t

Now, let's move on to part (b) to determine how long the ball is in the air:

(b) How long is the ball in the air?

To find this, we need to find the time when the ball hits the ground. The ball hits the ground when the vertical position (y) is equal to 0. Set the equation y = 0 and solve for t:
0 = 6 + 50t - 16t²

To solve this quadratic equation, we can either factor it or use the quadratic formula. Factoring it would require trial and error, so let's use the quadratic formula directly:
t = (-b ± √(b² - 4ac)) / 2a

In this case, a = -16, b = 50, and c = 6. Substituting these values into the formula, we get:
t = (-50 ± √(50² - 4(-16)(6))) / (2(-16))

Simplifying this expression will give us the time it takes for the ball to hit the ground.

Moving on to part (c):

(c) When is the ball at its maximum height?

The ball reaches its maximum height at the highest point of its trajectory. At this point, the vertical velocity (y') is equal to 0. Set the equation y' = 0 and solve for t:
0 = 50 - 32t

Solving this equation will give us the time it takes for the ball to reach its maximum height.

Finally, for part (d):

(d) Simulate the motion of the ball by graphing the equations found in part (a).

To graph the motion, plot the graph of y as a function of t using the equation found in part (a). This will give you the trajectory of the ball.

Read this over and you should be able to solve your problem.

The acceleration of a body is defined as the change of its velocity during an interval of time divided by the interval of time. Expressed algebraically, a = (Vo - Vf)/t where a = the acceleration, Vo = the initial velocity, Vf = the final velocity and t = the interval of time.

Uniformly Accelerated Motion

Three equations present the relationships between the initial velocity, the final velocity, the distance covered, acceleration and time.
The first expresses the relationship between the initial and final velocities and a time interval.
From the basic a = (Vo - Vf) we derive the first as
......................................t
..............................Vf = Vo + at (the acceleration is assumed constant)

The second defines the relationship between the distance traveled by a body exposed to a constant acceleration. The distance traveled over an interval of time derives from the product of the average velocity during the time interval and the time interval. The average velocity is (Vo + Vf)/2 and the time interval "t" which yields d = (Vo + Vf)t/2. Substituting the Vf derived above, d = [Vo + (Vo + at)]t/2 resulting in

...............................d = Vo(t) + a(t^2) (quite often, s is used in place of d)
......................................................2

The third combines the first two by eliminating the time interval "t".
Therefore, a(d) = (Vo - Vf)(Vo + Vf)t = (1/2)(Vf - Vo)(Vf + Vo) resulting in
.....................................t.............2

..............................Vf^2 = Vo^2 + 2ad

As written, these three equations apply to the uniformly accelerated motion of a body in a horizontal plane.

Another application of these equations is to falling bodies under the influence of the earth's gravity. All bodies fall to the ground due to the force of gravity pulling on them. As strange as it may seem, all bodies fall with the same acceleration, regardless of their masses. The acceleration of a falling body, caused by the force of gravity, depends on the distance of the body from the center of the earth. The gravitational force, and thereby the acceleration, decreases the further the body is from the center of the earth. The acceleration due to gravity is given the notaion "g" and averages 32.17 ft./sec.^2. Quite often, for solving instructional problems, g = 32 ft./sec.^2 is acceptable.

The equations of accelerated motion derived earlier also apply to falling (or rising) bodies with the exception that the term "a" for acceleration is replaced by the term "g). This results in

..............................Vf = Vo + gt (the acceleration is assumed constant)
...............................d = Vo(t) + g(t^2)
.....................................................2
..............................Vf^2 = Vo^2 + 2gd

As written, these expressions apply to falling bodies. The equations that apply to rising bodies are

..............................Vf = Vo - gt (the acceleration is assumed constant)
...............................d = Vo(t) - g(t^2)
.....................................................2
..............................Vf^2 = Vo^2 - 2gd

All of the above ignores surface and atmospheric friction.