Question Statement
The height of a projectile shot from the ground level is given by the position function:
S(t)=−16t2+256t
where S is measured in feet and t in seconds.
(i) Determine the height of the projectile at t=2, t=6, t=9, and t=10.
(ii) What is the average velocity of the projectile between t=2 and t=5?
(iii) Show that the average velocity between t=7 and t=9 is zero and interpret this result physically.
(iv) At what time does the projectile hit the ground?
(v) Determine the instantaneous velocity at time t=8.
(vi) What is the maximum height that the projectile attains?
Background and Explanation
This problem explores the kinematics of a projectile using a quadratic position function. To solve these parts, we use the concepts of function evaluation for height, the difference quotient for average velocity, and the limit definition of a derivative for instantaneous velocity.
Solution
To find the height at specific times, we substitute the values of t into the position function S(t)=−16t2+256t.
At t=2:
S(2)S(2)=−16(2)2+256(2)=−16(4)+512=−64+512=448 ft
At t=6:
S(6)S(6)=−16(6)2+256(6)=−16(36)+1536=−576+1536=960 ft
At t=9:
S(9)S(9)=−16(9)2+256(9)=−16(81)+2304=−1296+2304=1008 ft
At t=10:
S(10)S(10)=−16(10)2+256(10)=−1600+2560=960 ft
The average velocity is the change in position divided by the change in time: Vavg=t2−t1S(t2)−S(t1).
For t=2 and t=5:
VavgVavg=5−2S(5)−S(2)=3[−16(5)2+256(5)]−448=3[−400+1280]−448=3880−448=3432=144 ft/sec
To show the average velocity is zero between t=7 and t=9:
VavgVavg=9−7S(9)−S(7)=21008−[−16(7)2+256(7)]=21008−(−784+1792)=21008−1008=0 ft/sec
Physical Interpretation:
Physically, an average velocity of zero means the net displacement is zero. This indicates that the projectile reached its maximum height and returned to the exact same height (1008 ft) during this interval. The peak of the trajectory occurs exactly in the middle of these two times (at t=8).
The projectile hits the ground when the height S(t)=0.
−16t2+256t−16t(t−16)=0=0
Setting each factor to zero:
−16t=0⇒t=0 (This is the launch time)
t−16=0⇒t=16
The projectile hits the ground at t=16 seconds.
To find the instantaneous velocity, we use the limit definition of the derivative:
V(t1)V(t1)=Δt→0limΔtS(t1+Δt)−S(t1)=Δt→0limΔt[−16(t1+Δt)2+256(t1+Δt)]−[−16t12+256t1]=Δt→0limΔt−16(t12+Δt2+2t1Δt)+256t1+256Δt+16t12−256t1=Δt→0limΔt−16t12−16Δt2−32t1Δt+256Δt+16t12=Δt→0limΔtΔt(−16Δt−32t1+256)=−16(0)−32t1+256=−32t1+256
At t=8:
V(8)=−32(8)+256=−256+256=0 ft/sec
The maximum height is attained when the instantaneous velocity is zero.
From Part (v), we found V(t)=−32t+256.
−32t+25632tt=0=256=32256=8 sec
Now, substitute t=8 into the position function to find the maximum height:
S(8)S(8)=−16(8)2+256(8)=−16(64)+2048=−1024+2048=1024 ft
- Position Function: S(t)
- Average Velocity: Vavg=t2−t1S(t2)−S(t1)
- Instantaneous Velocity (Derivative): V(t)=limΔt→0ΔtS(t+Δt)−S(t)
- Ground Impact: Set S(t)=0 and solve for t.
- Maximum Height: Occurs when V(t)=0.
Summary of Steps
- Evaluate Heights: Plug specific time values into S(t) to find the position at those moments.
- Calculate Average Velocity: Use the slope formula (change in position over change in time) for the given intervals.
- Interpret Zero Velocity: Recognize that zero average velocity implies the object returned to its starting height within that interval.
- Find Ground Time: Solve the quadratic equation S(t)=0 using factoring.
- Derive Velocity Function: Use the limit definition of the derivative to find the general formula for velocity, V(t).
- Determine Max Height: Find the time when V(t)=0 and substitute that time back into the height formula S(t).