Applications of Parametric Equations

Question

Sanjay attempts a 49-yard field goal in a football game. For his attempt to be a success, the football needs to pass through the uprights and over the crossbar that is 10 feet above the ground.

Sanjay kicks the ball from the ground with an initial velocity of 73 feet per second, at an angle of 34° with the horizontal.

What is true of Sanjay's attempt?

 

 

a) The kick is good! The football clears the crossbar by approximately 5 feet.

b) The kick is not successful. The ball is approximately 2 feet too low.

c) The kick is not successful. The ball is approximately 5 feet too low.

d) The kick is not successful. The ball is approximately 8 feet too low.

Answer 1

We can break the initial velocity up into two components, the velocity in the x-direction (v_i-x) and the velocity in the y-direction (v_i-y), using trig ratios.

Because cos(θ) = adjacent/hypotenuse, V_i-x = 78cos(34°) = 64.665 ft/s

Because sin(θ) = opposite/hypotenuse, V_i-y = 78sin(34°) = 43.617 ft/s

In the y-direction, we can calculate the time it takes for the ball to get to the 10 ft height using:

y = v_i-yt + 1/2at² where y = the height above the ground, V_i-y is the initial velocity in the y-direction, t is the time the ball is in the air, and "a" is the acceleration of gravity in the y-direction (which is -32.17 ft/s²)

So:

10 = 43.617t + 1/2(-32.17)t²

16.085t² - 43.617t + 10 = 0

t = [-(-43.617) ± √(43.617² -4(16.085)(10))]/(2•16.085)

t = (43.617 + √1259)/32.17

t = 2.4588 s and t = 0.2528 s

The first time is on the left side of the field when the ball passes 10 ft high. The second time is the one we are interested in when the ball is "crossing the goal line"

Now, let's look at the x-direction to see how far the ball traveled in the 2.4588 seconds.

x = v_i-xt where x is the distance traveled in the x-direction, v_i-x is the initial velocity in the x-direction, and t is the time the ball is in the air (the same as the time the ball is in the air in the y-direction)

x = (64.665)(2.4588) = 159 ft = 53 yds. The ball sailed way over the goal post which was at 49 yds.

Answer 2

For this question we can assume that the the horizontal velocity is constant and therefore we find how much time it takes for the ball to travel 49 yards (147 ft). Our horizontal velocity is going to be the initial velocity multiplied by the cosine of the kick angle. So time is 147ft/(v_x) where v_x is v*cos(θ). So now that we know the time, we can find out what the height of the ball at the time it reaches the uprights. We calculate this height using one of the kinematic equations. We know the initial velocity in the vertical direction by multiplying initial velocity by sin(θ), from there we plug our initial vertical velocity and time into the following kinematic equation.

d=v_i*t+(1/2)*a*t²

If the ball is 10+ ft above the ground, then the kick is good.