How can quadratic functions be used to model projectile motion?

Question

Consider the following example below:

A soccer player kicks a ball into the air from a height of 2 meters with an initial vertical velocity of 15 m/s. The height of the ball after t seconds is given by the quadratic function:

h(t) = -4.9t² + 15t +2

where h(t) represents the height of the ball in meters at time t seconds.

a) How long does it take for the ball to reach its maximum height?

b) What is the maximum height the ball reaches?

c) How long is the ball in the air before it hits the ground?

d) At what time(s) is the ball at a height of 10 meters?

Answer 1

For a) and b) transform h(t) into vertext form.

For c) set h(t) = 0 and solve for t

For d) set h(t) = 10 and solve for t

Answer 2

The projectile reaches it maximum height of 13 .48 meters in 1.53 seconds

The projectile is back on the ground after 3.189 seconds

During its upward part of the path the projectile reaches a height 10 meters in 0.688 seconds

On the way back down the projectile falls back down to 10 meters after 2.373 seconds

The flight path of a projectile, typically forms a parabolic trajectory. This path is influenced by factors such as the initial velocity, launch angle, and air resistance.

Gravity is main force working against a projectile’s initial upward velocity. Initially the upward velocity overcomes gravity. However, since gravity is constantly pulling down against the projectile during its flight, this results in a curved path. At the highest point of its travel the projectile no longer has an upward velocity to overcome gravity and speeds downward.

Flight path of a projectile has the shape of a U, an upside down U a parabola that opens downward. Quadratic Equations are used as mathematical representations of a squared variable and scenarios involving curved paths; parabolas

For your example

h(t) = -4.9t² + 15t +2

You can line this up with the standard form of a Quadratic, f(x) = ax² + bx + c

h(t) = -4.9t² + 15t +2

f(x) = -4.9x² + 15x + 2

a = -4.9, b = 15 and c = 2

The maximum height can be found by calculating the vertex h the x component of the vertex = -b/2a; the x component represents the time it takes for the projectile to reach maximum height

-b/2a = (-15)/(2*-4.9)

(-15/-9.8) = 1.53 = h

The y or k component of the vertex can be found by plugging in the x component; the y component of the vertex is the maximum height

f(1.53) = -4.9(1.53)² + 15(1.53) + 2

f(1.53) = -4.9(2.34) + 22.95 + 2

f(1.53) = -11.47 + 22.95 + 2

f(1.53) = 13.48 = y

The vertex is (1.53, 13.48)

The ball reaches a maximum height of 13.48 meters in 1.53 seconds

The height is 0 when the projectile returns to the ground, and remember time is positive

0 = -4.9x + 15x + 2

You can use the Quadratic formula to solve for x when f(x) is

x = -b[+/-((b² - 4ac))^1/2]/2a

x = {-15[+/-((225 + 39.2))^1/2]}/-9.8

x = {-15+/- (√264.2)}/-9.8

x = (-15 - 16.25)/-9.8

x = -31.25/-9.8

x = 3.189 seconds

The projectile is in the air for 3.189 seconds

10 = -4.9x² + 15x + 2

Subtract 10 from both sides of the equation then use the Quadratic formula again to yield a height of 10 meters after 0.688 seconds upward on the flight path and 2.373 seconds downward along the paths.

The methods I used above were algebraic, if you are familiar with derivatives you can use those as well.

By the way you can graph the function and you should to confirm the numbers above.

I hope this helps.

Answer 3

The first derivative of a quadratic in standard form is 2ax + b. The minimum or maximum occurs when this value (the slope of the quadratic in general form) is zero. So set this derivative to zero and solve for x: 2ax+b = 0, 2ax = -b, x= -b/(2a). Note that this is the same x value as given before.

By the way, this is the x-value of the vertex. Converting to vertex form is another way of finding this value.