The trajectory of a golf ball hit from a tee on the ground at an angle of 50 degrees with an initial speed of 60 meters per second can be modeled by the parabol

Question

 

The trajectory of a golf ball hit from a tee on the ground at an angle of 50 degrees with an initial speed of 60 meters per second can be modeled by the parabola f(x) = 1.19x − 0.0033x2, where the x-axis is the ground. Find the height of the highest point of the trajectory and the horizontal distance the golf ball travels before hitting the ground.

 

Mostly just need to know how to do it.

 

Answer 1

x is apparently the distance on the ground from the point where the ball is hit.

 

the horizontal speed of the ball can be given by s and can be derived from equation cos(50) = s/60; s = 38.5 mps

 

f(x) represents height. So, the parabola's highest point would be at -b/2a = -1.19/-.0066 = 180.3 meters, or that ball has traveled 180.3 meters

 

plug this into formula to get a height of 107.28 meters.

 

To find distance traveled note that ball has to be on the ground so f(x) = 0

 

f(x) = 0, when x = 0, that makes sense.

 

But when is f(x) next equal to zero?

 

-.0033x^2 + 1.19x = 0

 

x*(-.0033x + 1.19) = 0

 

x= 1.19/.0033 = 360 .6 meters

 

What's interesting about the problem is that we did not use the speed of the ball. Why is that? Because, once the formula for f(x) has been determined, the speed is already incorporated into that formula. If the ball had been hit at a different speed, we would have a different formula. If the ball had been hit a 1 mps, obviously the f(x) formula would not describe its path.

 

Now, if it had been asked how long did the ball travel in the air, now we would use the speed. it would be 360.6/38.5 = 9.35 sec. That time seems reasonable.