Question 3.1.5: Solving a Problem Involving Projectile Motion A ball is proj...
Solving a Problem Involving Projectile Motion
A ball is projected directly upward from an initial height of 100 ft with an initial velocity of 80 ft per sec.
(a) Give the function that describes the height of the ball in terms of time t.
(b) After how many seconds does the ball reach its maximum height? What is this maximum height?
(c) For what interval of time is the height of the ball greater than 160 ft?
(d) After how many seconds will the ball hit the ground?
ALGEBRAIC SOLUTION
(a) Use the projectile height function.
s(t) =-16t²+ν_{0}t+s_{0} \text{Let} ν_{0}=80 and
s(t) =-16t² + 80t + 100 s_{0}=100.
(b) The coefficient of t² is -16, so the graph of the projectile function is a parabola that opens down. Find the coordinates of the vertex to determine the maximum height and when it occurs. Let a = -16 and b = 80 in the vertex formula.
t=-\frac{b}{2a}=-\frac{80}{2(-16)}=2.5
s(t)= – 16t²+80t+100
s(2.5)= – 16(2.5)²+80(2.5)+100
s(2.5)=200
Therefore, after 2.5 sec the ball reaches its maximum height of 200 ft.
(c) We must solve the related quadratic inequality.
-16t² + 80t + 100 > 160
-16t² + 80t – 60 >0 Subtract 160.
4t² – 20t + 15 < 0 Divide by -4. Reverse the
inequality symbol.
Use the quadratic formula to find the solutions of 4t² – 20t + 15 = 0.
t=\frac{-(-20)\pm \sqrt{(-20)^{2}-4(4)(15)} }{2(4)} Here a = 4, b = -20, and c = 15.
t=\frac{5-\sqrt{10} }{2} \approx 0.92 \text{or} t=\frac{5+\sqrt{10} }{2}\approx 4.08
These numbers divide the number line into three intervals:
(-∞, 0.92), (0.92, 4.08), and (4.08, ∞).
Using a test value from each interval shows that (0.92, 4.08) satisfies the inequality. The ball is greater than 160 ft above the ground between 0.92 sec and 4.08 sec.
(d) The height is 0 when the ball hits the ground. We use the quadratic formula to find the positive solution of the equation
-16t² + 80t + 100 = 0.
Here, a = -16, b = 80, and c = 100.
t=\frac{-80\pm \sqrt{80^{2}-4(-16)(100)} }{2(-16)}
t≈-\underset{\text{Reject}}{\xcancel{ 1.04}} \text{or} t≈6.04
The ball hits the ground after about 6.04 sec.
GRAPHING CALCULATOR SOLUTION
(a) Use the projectile height function as in the algebraic solution, with ν_{0} = 80 \text{and} s_{0} = 100.
s(t) = – 16t²+80t+100
(b) Using the capabilities of a calculator, we see in Figure 11 that the vertex coordinates are indeed (2.5, 200).
(c) If we graph
y_{1} = -16x² + 80x + 100 \text{ and} y_{2} = 160,
as shown in Figures 12 and 13, and locate the two points of intersection, we find that the x-coordinates for these points are approximately
0.92 and 4.08.
Therefore, between 0.92 sec and 4.08 sec, y_{1} is greater than y_{2} , and the ball is greater than 160 ft above the ground.
(d) Figure 14 shows that the x-intercept of the graph of y = -16x² + 80x + 100 in the given window is approximately (6.04, 0), which means that the ball hits the ground after about 6.04 sec.
