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
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