Projectile Motion Formulas
f(x) = a𝒙 𝟐
+ bx + c
When objects are thrown.
feet:
𝒉 𝒕 = −𝟏𝟔𝒕 𝟐
+ 𝒗 𝟎 𝒕 + 𝒉 𝟎
When objects are dropped.
feet:
𝒉 𝒕 = −𝟏𝟔𝒕 𝟐 + 𝒉 𝟎
meters:
𝒉 𝒕 = −𝟒. 𝟗𝒕 𝟐
+ 𝒗 𝟎 𝒕 + 𝒉 𝟎
meters:
𝒉 𝒕 = −𝟒. 𝟗𝒕 𝟐
+ 𝒉 𝟎
Projectile Motion Critical Feature Formulas
roots
𝒙 =
−𝒃 ± 𝒃 𝟐 − 𝟒𝒂𝒄
𝟐𝒂
y-intercept
(0, y)
vertex
−𝒃
𝟐𝒂
, 𝒇
−𝒃
𝟐𝒂
,
Initial height of object.
x: time it takes to reach max height
y: max height
Time the object is on the ground.
Only + values.height time
12
9
6
3
1 2 3 4
height(feet)
time (seconds)
The y-intercept is the height in which the
object begins. It is when x, seconds, is zero.
In this example the y-intercept
is (0, 6) The basketball begins
at 6 feet. This is the “c” value
of the quadratic equation.
𝒉 𝒕 = 𝒂𝒕 𝟐
+ 𝒃𝒕 + 𝒄
𝒉 𝒕 = 𝒂𝒕 𝟐
+ 𝒃𝒕 + 𝟔
12
9
6
3
1 2 3 4
height(feet)
time (seconds)
The vertex gives the amount of time, x in
seconds, the object reaches its maximum height,
y in feet or meters.
In this example the vertex is
(2, 12). In 2 seconds the basket-
ball reaches its maximum height
of 12 feet.
Use the vertex formula to find
the maximum height and time.
𝒉 𝒕 = 𝒂𝒕 𝟐
+ 𝒃𝒕 + 𝒄 →
−𝒃
𝟐𝒂
, 𝒇
−𝒃
𝟐𝒂
,
12
9
6
3
1 2 3 4
height(feet)
time (seconds)
The positive root gives the amount of time, x in
seconds, it takes the object to hit the ground,
when y is zero feet or meters.
In this example the positive root
is (4, 0). In 4 seconds the basket-
ball reaches the ground of zero
feet.
Use the quadratic formula or
factor the quadratic to find
how long it takes the object
to reach the ground.
𝒉 𝒕 = 𝒂𝒕 𝟐
+ 𝒃𝒕 + 𝒄 → 𝒙 =
−𝒃± 𝒃 𝟐−𝟒𝒂𝒄
𝟐𝒂

Projectile motion overview

  • 1.
    Projectile Motion Formulas f(x)= a𝒙 𝟐 + bx + c When objects are thrown. feet: 𝒉 𝒕 = −𝟏𝟔𝒕 𝟐 + 𝒗 𝟎 𝒕 + 𝒉 𝟎 When objects are dropped. feet: 𝒉 𝒕 = −𝟏𝟔𝒕 𝟐 + 𝒉 𝟎 meters: 𝒉 𝒕 = −𝟒. 𝟗𝒕 𝟐 + 𝒗 𝟎 𝒕 + 𝒉 𝟎 meters: 𝒉 𝒕 = −𝟒. 𝟗𝒕 𝟐 + 𝒉 𝟎
  • 2.
    Projectile Motion CriticalFeature Formulas roots 𝒙 = −𝒃 ± 𝒃 𝟐 − 𝟒𝒂𝒄 𝟐𝒂 y-intercept (0, y) vertex −𝒃 𝟐𝒂 , 𝒇 −𝒃 𝟐𝒂 , Initial height of object. x: time it takes to reach max height y: max height Time the object is on the ground. Only + values.height time
  • 3.
    12 9 6 3 1 2 34 height(feet) time (seconds) The y-intercept is the height in which the object begins. It is when x, seconds, is zero. In this example the y-intercept is (0, 6) The basketball begins at 6 feet. This is the “c” value of the quadratic equation. 𝒉 𝒕 = 𝒂𝒕 𝟐 + 𝒃𝒕 + 𝒄 𝒉 𝒕 = 𝒂𝒕 𝟐 + 𝒃𝒕 + 𝟔
  • 4.
    12 9 6 3 1 2 34 height(feet) time (seconds) The vertex gives the amount of time, x in seconds, the object reaches its maximum height, y in feet or meters. In this example the vertex is (2, 12). In 2 seconds the basket- ball reaches its maximum height of 12 feet. Use the vertex formula to find the maximum height and time. 𝒉 𝒕 = 𝒂𝒕 𝟐 + 𝒃𝒕 + 𝒄 → −𝒃 𝟐𝒂 , 𝒇 −𝒃 𝟐𝒂 ,
  • 5.
    12 9 6 3 1 2 34 height(feet) time (seconds) The positive root gives the amount of time, x in seconds, it takes the object to hit the ground, when y is zero feet or meters. In this example the positive root is (4, 0). In 4 seconds the basket- ball reaches the ground of zero feet. Use the quadratic formula or factor the quadratic to find how long it takes the object to reach the ground. 𝒉 𝒕 = 𝒂𝒕 𝟐 + 𝒃𝒕 + 𝒄 → 𝒙 = −𝒃± 𝒃 𝟐−𝟒𝒂𝒄 𝟐𝒂