a. When will the projectile's height above ground be 768 ft?
b. When will the projectile's height above ground be at least 768 ft?
c. When will the projectile's height above ground be less than or equal to 768 ft?
A projectile is launched straight up from ground level with an initial velocity of 256 ft/sec.?
2015-09-04 2:59 am
回答 (1)
2015-09-04 5:33 am
Take g = 32 ft/s²
Take upward quantities as positive.
(a)
When s = 768 (m)
vₒt + (1/2)at² = 768
(256)t + (1/2)(-32)t² = 768
256t -16 t² = 768
t² - 16t + 48 = 0
(t - 4)(t - 12) = 0
t = 4 or t = 12
i.e. time taken = 4 s or 12 s
(b)
s ≥ 768
vₒt + (1/2)at² ≥ 768
(256)t + (1/2)(-32)t² ≥ 768
256t -16 t² ≥ 768
16t - t² ≥ 48
t² - 16t + 48 ≤ 0
(t - 4)(t - 12) ≤ 0
4 ≤ t ≤ 12
i.e. 4 s ≤ time taken ≤ 12 s
(c)
s ≤ 768
vₒt + (1/2)at² ≤ 768
(256)t + (1/2)(-32)t² ≤ 768
256t -16 t² ≤ 768
16t - t² ≤ 48
t² - 16t + 48 ≤ 0
(t - 4)(t - 12) ≤ 0
t ≤ 4 or t ≥ 12
i.e. time taken ≤ 4 s or time taken ≥ 12 s
Take upward quantities as positive.
(a)
When s = 768 (m)
vₒt + (1/2)at² = 768
(256)t + (1/2)(-32)t² = 768
256t -16 t² = 768
t² - 16t + 48 = 0
(t - 4)(t - 12) = 0
t = 4 or t = 12
i.e. time taken = 4 s or 12 s
(b)
s ≥ 768
vₒt + (1/2)at² ≥ 768
(256)t + (1/2)(-32)t² ≥ 768
256t -16 t² ≥ 768
16t - t² ≥ 48
t² - 16t + 48 ≤ 0
(t - 4)(t - 12) ≤ 0
4 ≤ t ≤ 12
i.e. 4 s ≤ time taken ≤ 12 s
(c)
s ≤ 768
vₒt + (1/2)at² ≤ 768
(256)t + (1/2)(-32)t² ≤ 768
256t -16 t² ≤ 768
16t - t² ≤ 48
t² - 16t + 48 ≤ 0
(t - 4)(t - 12) ≤ 0
t ≤ 4 or t ≥ 12
i.e. time taken ≤ 4 s or time taken ≥ 12 s
收錄日期: 2021-04-18 00:08:44
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