Question Statement A projectile is launched vertically upward from an initial height of 129 feet with an initial velocity of 87
(i) What are the position, velocity, and acceleration functions?
(ii) When will the projectile hit the ground?
(iii) What is its impact velocity?
(iv) When will the projectile reach its maximum height?
(v) What is the maximum height of projectile?
Background and Explanation This problem applies one-dimensional kinematics with constant acceleration due to gravity. By integrating the constant acceleration function and applying initial conditions, we derive the velocity and position functions needed to analyze the projectile's motion.
Solution Given the initial conditions:
Initial height: S 0 = 129
Initial velocity: v i = 87
Acceleration due to gravity: g = 32 2
Since upward is defined as positive, the acceleration is constant and negative:
a = − 32 ft/s 2
To find the velocity function, integrate acceleration with respect to time:
v = ∫ a d t = − ∫ 32 d t = − 32 t + A
Apply the initial condition at t = 0 v = 87 87 = − 32 ( 0 ) + A ⇒ A = 87
Thus, the velocity function is:
v = 87 − 32 t ft/s
To find the position function, integrate the velocity:
S = ∫ v d t = ∫ ( 87 − 32 t ) d t = 87 t − 2 1 ( 32 ) t 2 + B = 87 t − 16 t 2 + B
Apply the initial condition at t = 0 S = 129 129 = 87 ( 0 ) − 16 ( 0 ) 2 + B ⇒ B = 129
Therefore, the position function is:
S = 129 + 87 t − 16 t 2 ft
The projectile hits the ground when S = 0 129 + 87 t − 16 t 2 = 0
Rearranging into standard quadratic form:
16 t 2 − 87 t − 129 = 0
Using the quadratic formula t = 2 a − b ± b 2 − 4 a c t = 2 ( 16 ) − ( − 87 ) ± ( − 87 ) 2 − 4 ( 16 ) ( − 129 ) t = 32 87 ± 7569 + 8256 t = 32 87 ± 15825
Calculating the two roots:
t = 32 87 + 125.8 or t = 32 87 − 125.8 t = 32 212.8 ≈ 6.65 s or t = 32 − 38.8 ≈ − 1.21 s
Since time must be positive, the projectile hits the ground at t = 6.65
Using the time of impact calculated above:
Time of impact = 6.65 − 2.71875 = 3.93125 s
Substituting into the velocity equation:
v = 87 + 32 t v = 87 + 32 ( 3.93125 ) v = 87 + 125.8 v = 212.2 ft/s
At maximum height, the velocity is momentarily zero:
v = 87 − 32 t = 0 − 32 t = − 87 t = 32 87 = 2.71875 s
Substitute t = 2.71875 S = 129 + 87 ( 2.71875 ) − 16 ( 2.71875 ) 2 S = 247.265625 ft
Constant acceleration: a = − g = − 32 2 Velocity from acceleration: v ( t ) = ∫ a d t = v 0 + a t Position from velocity: S ( t ) = ∫ v d t = S 0 + v 0 t + 2 1 a t 2 Quadratic formula: t = 2 a − b ± b 2 − 4 a c a t 2 + b t + c = 0 Maximum height condition: v = 0
Summary of Steps
Establish acceleration a = − 32 2 v ( t ) = 87 − 32 t Integrate velocity to get position S ( t ) = 129 + 87 t − 16 t 2 S 0 = 129 Solve for impact time by setting S ( t ) = 0 t ≈ 6.65 Calculate impact velocity by substituting the time of impact into the velocity functionFind time to maximum height by setting v ( t ) = 0 t = 2.71875 Compute maximum height by substituting t = 2.71875