During a recent eruption, a volcano spewed copious amounts of ash. One small piece of ash was ejected from the volcano with an initial velocity of 368 ft/sec. The height H, in feet, of our ash projectile is given by the equation:H = - 16t^2 + 368t

Question 1. When does the ash projectile reach the its maximum height? t=?

Question 2. What is its maximum height?

Question 3. When does the ash projectile return to the ground?

Question

Douglas Mccormick

Mathematics

24 June, 10:29

During a recent eruption, a volcano spewed copious amounts of ash. One small piece of ash was ejected from the volcano with an initial velocity of 368 ft/sec. The height H, in feet, of our ash projectile is given by the equation:H = - 16t^2 + 368t

Question 1. When does the ash projectile reach the its maximum height? t=?

Question 2. What is its maximum height?

Question 3. When does the ash projectile return to the ground?

+1

Answers (1)

Know the Answer?

Cassius Hatfield · Answer 1

The formula is:

H = - 16 t² + 368 t

The projectile reaches its maximum at:

t max = - b / 2 a = - 368 / ( - 32) = 11.5 s

- 16 t² + 368 t = 0

t ( - 16 t + 368) = 0

t 1 = 0 s, t 2 = 368 : 16 = 23 s

H max = - 16 · 11.5 + 368 · 11.5 = - 2116 + 4232 = 2116 ft

Answers:

1. The projectile reaches its maximum height at t = 11.5 s.

2. H max = 2116 ft.

3. The projectile returns to the ground after t = 23 s.