Parametric Word Problem

(e)  Simultaneously graph the equations found in part (a).

 

http://www.wolframalpha.com/input/?i=x%3D(145cos20)t+,++++++y+%3D-16t%5E2%2B(145sin20)t

 

You'd do it just by plotting the information that you have found :)

Ichiro throws a baseball with an initial speed of 145 feet per second at an angle of 20° to the horizontal.  The ball leaves Ichiro's hand at a height of 5 feet.

I am going to let gravity be 32 feet/sec^2

I am going to put this on a grid where the hand is at at the point (0,0)

So the ball will het the ground where y=-5

 

time is in seconds

Acceleration is in feet/sec^2

Velocity is in feet/sec

Displacement is in feet

 

(a)  Find parametric equations that model the position of the ball as a function of time.

Initial condition:

\(\ddot x=0\\ \dot x =145cos20\\ x=0\\~\\ \ddot y = -32\;feet/sec^2\\ \dot y = 145sin20\\ y =0 \\~\\ \)

\(Continuing\; equations:\\ \ddot x=0\\ \dot x =145cos20\\ x=(145cos20)t\\~\\ \ddot y = -32\\ \dot y = -32t+(145sin20)\\ y =-16t^2+(145sin20)t\\~\\ \mbox{So the parametric model of the position of the ball is}\\ x=(145cos20)t \qquad and \qquad y =-16t^2+(145sin20)t\)

Ichiro throws a baseball with an initial speed of 145 feet per second at an angle of 20° to the horizontal.  The ball leaves Ichiro's hand at a height of 5 feet.

 

I am going to let gravity be 32 feet/sec^2

I am going to put this on a grid where the hand is at at the point (0,0)

So the ball will be on the ground where y=-5

Acceleration is in feet/sec^2

Velocity is in feet/sec

Displacement is in feet

 

 

\( \mbox{The parametric model of the position of the ball is}\\ x=(145cos20)t \qquad and \qquad y =-16t^2+(145sin20)t\)

 

(b)  How long is the ball in the air?

The ball will be in the air until y=-5

\(-5=-16t^2+(145sin20)t\\ -16t^2+(145sin20)t+5=0\\ 16t^2-(145sin20)t-5=0\\ 16t^2-49.59292078227t-5=0 \qquad approx\\\)

 

145*sin(pi/9) = 49.59292078227

(49.59292078227+sqrt(49.59292078227^2+320))/32 = 3.1972963766958047

 

After approximately 3.2 seconds the ball will his the ground.

 

 

 

 

 

(c)  Determine the horizontal distance that the ball travels.

When t=3.1972963766958047 seconds find x

 

x=(145cos20)*3.1972963766958047

x=(145*cos(pi/9))*3.1972963766958047

(145*cos(pi/9))*3.1972963766958047 = 435.648992688795078977

The ball travels  approx 436 feet in the horizonal direction.

 

 

 

 

(d)  When is the ball at its maximum height?  Determine the maxium height of the ball.

The ball will be at its maximum height when dy/dt = 0

 

\(-32t+(145*sin(20))=0\\ 32t=+(145*sin(20))\\ t=145*sin(\pi/9)\div32\qquad\\ \mbox{I have changed to radians because the forum calc works in radians}\\\)

145*sin(pi/9)/32 = 1.5497787744459375

So the ball is at its highest after approximately 1.55 seconds

 

\(y=-16t^2+(145*sin20)t \)

t=1.5497787744459375 seconds

y=-16*1.5497787744459375^2+145*sin(pi/9)*1.5497787744459375 = 38.4290279955704323242

So the ball goes up 38.4 feet above the hand BUT the hand is 5 feet above the ground

So

 

The ball reaches a maximum height of  approx 38.4+5 = 43.4 feet

Thanks for the points Gibsonj338  I appreciate that

BUT

Do you understand my answer?  Do you hav any questions ?