A baseball player throws baseball at a velocity 139ft per second at an angle of 20 degrees. The ball leaves the players hand at a height of 5 feet. The path of the ball has a parametric equation x=v(o)t cos(o) and y-h+v(o)tsin(o)-16t^2. Write the parametric equations that describe the motion of the ball as a function of time. How long does it take for the ball to hit the ground? When is the ball at its maximum height? What is the maximum height of the ball?
typo
you mean y=h+v(o)tsin(o)-16t^2
Vi = initial speed up = 139 sin 20
= 47.54 ft/s
u = horizontal speed = 139 cos 20
= 131 ft/s
x = 131 t
y = 5 + 47.54 t - 16 t^2
when is y = 0?
16 t^2 - 47.54 t - 5 = 0
t = [ 47.54 +/- sqrt(2260+320)]/32
use positive time
t = 3.07 seconds in the air
Now I suppose you could use completing the square to find the vertex of that parabola giving the max height and time thereof. However it is easier to use either physics or calculus
when is the up velocity = 0? That is the top.
dy/dt = up velocity
= 0 = 47.54 - 32 t
= 1.49 seconds to stop at top
then y at top =
5+47.54(1.49)-16(1.49)^2
= 40.3 ft
by the way surely they ask the range, x
x = u * t in air
= 131 * 3.07 = 402 ft
That is a serious throw
Given the parametric equations:
x = v₀t cos(θ)
y = h + v₀t sin(θ) - 16t²
Where:
v₀ = initial velocity of the ball = 139 ft/s
θ = launch angle = 20 degrees
h = initial height = 5 feet
t = time elapsed
First, let's convert the launch angle from degrees to radians:
θ_rad = (20 * π) / 180
θ_rad ≈ 0.3491 radians
Now, we can plug in the values into the parametric equations:
x = 139t cos(0.3491)
y = 5 + 139t sin(0.3491) - 16t²
To determine when the ball hits the ground, we set y = 0:
0 = 5 + 139t sin(0.3491) - 16t²
This is a quadratic equation in terms of t. By solving the equation, we can find the value(s) of t when the ball hits the ground. Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
Where a = -16, b = 139 sin(0.3491), and c = 5, we can solve for t.
To find the time when the ball is at its maximum height, we can determine when the vertical velocity is zero. Differentiate the y-equation with respect to t:
dy/dt = 139 sin(0.3491) - 32t
Set this expression equal to zero:
0 = 139 sin(0.3491) - 32t
Solving for t will give us the time when the ball reaches its maximum height.
Finally, to find the maximum height of the ball, substitute the time (t) obtained in the previous step into the y-equation.
Calculating these values will give us the complete information about the motion of the ball.
To write the parametric equations that describe the motion of the ball as a function of time, we can use the given equations:
x = v₀t cos(θ)
y = h + v₀t sin(θ) - 16t²
Where:
- x and y are the coordinates of the ball at a specific time t,
- v₀ is the initial velocity of the ball,
- θ is the launch angle of the ball,
- h is the initial height of the ball (5 feet in this case), and
- t is the time elapsed since the ball was thrown.
We are given that the initial velocity (v₀) is 139 ft/s and the launch angle (θ) is 20 degrees. Let's define these values in the equations:
x = 139t cos(20°)
y = 5 + 139t sin(20°) - 16t²
To find how long it takes for the ball to hit the ground, we need to determine the value of t when y = 0 (since y represents the vertical position). Setting the y equation equal to zero:
0 = 5 + 139t sin(20°) - 16t²
This is a quadratic equation that can be solved by factoring, completing the square, or using the quadratic formula. After finding the roots, we discard any negative values since time cannot be negative in this context. The positive root will give us the time it takes for the ball to hit the ground.
To find when the ball is at its maximum height, we can observe that at the highest point, the vertical velocity (dy/dt) will be zero. We can differentiate the y equation with respect to time (t) and solve for t when the derivative is zero.
To find the maximum height of the ball, we can plug in the time value obtained for the maximum height (found in the previous step) into the y equation.
By following these steps, we can answer each question precisely.