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The Parametric Equations for the Path of a Projectile Launched θ\theta

Question 39

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The parametric equations for the path of a projectile launched at a height h feet above the ground, at an angle θ\theta with the horizontal and having an initial velocity of v0v _ { 0 } feet per second is given by x=(v0cosθ) tx = \left( v _ { 0 } \cos \theta \right) t and y=h+(v0sinθ) t16t2y = h + \left( v _ { 0 } \sin \theta \right) t - 16 t ^ { 2 } . The center field fence in a ballpark is 10 feet high and 400 feet from home plate. The ball is hit 2 feet above the ground. It leaves the bat at an angle of θ\theta degrees with the horizontal at a speed of 100 miles per hour as shown in the figure. Write a set of parametric equations for the path of the ball.

 The parametric equations for the path of a projectile launched at a height h feet above the ground, at an angle  \theta  with the horizontal and having an initial velocity of  v _ { 0 }  feet per second is given by  x = \left( v _ { 0 } \cos \theta \right)  t  and  y = h + \left( v _ { 0 } \sin \theta \right)  t - 16 t ^ { 2 } . The center field fence in a ballpark is 10 feet high and 400 feet from home plate. The ball is hit 2 feet above the ground. It leaves the bat at an angle of  \theta  degrees with the horizontal at a speed of 100 miles per hour as shown in the figure. Write a set of parametric equations for the path of the ball.    A)  x = \left( \frac { 440 } { 3 } \cos \theta \right)  t , y = 2 + \left( \frac { 440 } { 3 } \sin \theta \right)  t  B)   x = 2 + \left( \frac { 440 } { 3 } \sin \theta \right)  t , y = \left( \frac { 440 } { 3 } \cos \theta \right)  t  C)  x = 2 + \left( \frac { 440 } { 3 } \sin \theta \right)  t - 16 t ^ { 2 } , y = \left( \frac { 440 } { 3 } \cos \theta \right)  t  D)   x = \left( \frac { 440 } { 3 } \sin \theta \right)  t , y = 2 + \left( \frac { 440 } { 3 } \cos \theta \right)  t - 16 t ^ { 2 }  E)  x = \left( \frac { 440 } { 3 } \cos \theta \right)  t , y = 2 + \left( \frac { 440 } { 3 } \sin \theta \right)  t - 16 t ^ { 2 }


A) x=(4403cosθ) t,y=2+(4403sinθ) tx = \left( \frac { 440 } { 3 } \cos \theta \right) t , y = 2 + \left( \frac { 440 } { 3 } \sin \theta \right) t
B) x=2+(4403sinθ) t,y=(4403cosθ) tx = 2 + \left( \frac { 440 } { 3 } \sin \theta \right) t , y = \left( \frac { 440 } { 3 } \cos \theta \right) t
C) x=2+(4403sinθ) t16t2,y=(4403cosθ) tx = 2 + \left( \frac { 440 } { 3 } \sin \theta \right) t - 16 t ^ { 2 } , y = \left( \frac { 440 } { 3 } \cos \theta \right) t
D) x=(4403sinθ) t,y=2+(4403cosθ) t16t2x = \left( \frac { 440 } { 3 } \sin \theta \right) t , y = 2 + \left( \frac { 440 } { 3 } \cos \theta \right) t - 16 t ^ { 2 }
E) x=(4403cosθ) t,y=2+(4403sinθ) t16t2x = \left( \frac { 440 } { 3 } \cos \theta \right) t , y = 2 + \left( \frac { 440 } { 3 } \sin \theta \right) t - 16 t ^ { 2 }

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