A toy rocket is shot vertically into the air from a 80-foot platform. The height

Question

A toy rocket is shot vertically into the air from a 80-foot platform. The height, h(t) in feet, of the rocket above the ground after t seconds is modeled by h(t) = -16t^2 + 128t + 80. (Give solutions accurate to two decimal places. No credit will be awarded for a calculator-only solution.) (a) How high is the rocket after 3 seconds? (b) How many seconds will it take the rocket to reach its maximum height? (c) What is the maximum height of the rocket? (d) How long will it take the rocket to reach the ground?

Explanation / Answer

(a). After 3 seconds, the height of the rocket will be h(3) = -16(3)2+128(3)+80 = -16*9+128*3+80 = -144 +384 +80= 320 ft.;

(b). We have h(t)= -16t2 +128t +80.Then dh/dt = -32t+128 and d2h/dt2= -32. Since d2h/dt2 is negative, therefore h(t) will be maximum when dh/dt = 0 or, -32t+128 = 0 or, t = 128/32 = 4. Thus, the rocket will take 4 seconds to achieve its maximum height.

( c ). We have h(t) = -16t2 +128t +80 = -16(t2 -8t) +80 = -16(t2 – 2t*4 + 16)+ 80+16*16 = -16(t-4)2+ 336. This is the equation of a parabola, with vertex at (4,336), which opens downwards. Since the vertex is the highest point of a downward opening parabola, hence the maximum height of the rocket is 336 ft.

(d). The rocket will reach the ground when h(t) = 0, i.e. when -16t2 +128t +80 = 0. On dividing both the sides by -16, we get t2 -8t -5 = 0. Now, on using the quadratic formula, we have t = [-(-8)±{(-8)2-4*1*(-5)}]/2*1 = [8±( 64+20)]/2 = (8±84)/2 = 4±21. Since t cannot be negative, we have t = 4+21 = 4 +4.58 = 8.58 (on rounding off to 2 decimal places). Thus, the rocket will reach the ground after 8.58 seconds.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.