A soccer ball kicked at the goal travels in a path given by the parametric equations: x=50t; #y=-16t^2+32t#, Suppose the ball enters the goal at a height of 5ft. How far away from the goal was the kicker?

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Answer 1 · 2017-04-09T11:49:00

The kicker was #50(1 pm sqrt(11)/4)# ft away from the goal.

Let us find time #t# when the ball is at a height of 5 ft.

By setting #y=5#,

#-16t^2+32t=5 Rightarrow -16t^2+32t-5=0#

By Quadratic Formula,

#t=1 pm sqrt(11)/4#

If the ball enters the goal on the way up, then the kicker was

#x=50(1 -sqrt(11)/4)# ft away from the goal.

If the ball enters the goal on the way down, then the kicker was

#x=50(1+sqrt(11)/4)# ft away from the goal.

I hope that this was clear.