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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?

Respuesta :

HomertheGenius
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.

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