How do write a parametric equation for this problem?

Question

Derek hits a ball with an initial velocity of 155 feet per second at an angle of 22 degrees above the horizontal. The ball travels straight at the 420 foot mark on the center field wall which is 15 feet high

Answer 1

We can consider the ball's motion in two different directions: vertical and horizontal. Air resistance is ignored.

For time t, the height of the ball is given by h(t)=u_Vt-½gt² where u_V is the vertical component of the initial velocity and g=32 ft/s² is the acceleration due to gravity pulling the ball to the ground.

The horizontal distance, d(t), is unaffected by gravity and d=u_Ht.

u_V=155sin(22)=58.06 ft/s approx.; u_H=155cos(22)=143.71 ft/s approx.

We now have two equations with t (time in seconds) as parameter.

h(t)=155tsin(22)-16t²; d(t)=155tcos(22).

The horizontal distance is 420 feet=155tcos(22), and h(t)=15 feet.

t=420/(155cos(22)) is the time it takes to reach the wall.

h(t)=155tsin(22)-16t². This has 2 solutions because the ball rises to a height of 15 feet in its trajectory before falling to the same height some time later. The relevant height of the ball at a distance of 420 feet from Derek is about 33 feet, that is, 18 feet above the wall, so the ball easily clears the wall.

The physical trajectory of the ball can be graphed by substituting for t, in terms of x, in h(t) so that we have h(x). From this it's easy to see the vertical and horizontal positions of the ball independent of time.