A projectile is launched straight up from ground level with an initial velocity of v0 per...
- A projectile is launched straight up from ground level with an initial velocity of v0 per sec. Neglecting air resistance, its height in feet, s, after t seconds after launch is given by the equation
s = –16t2 + v0t.
- At what time(s) will the projectile reach a height of 112 feet when v0 = 96 feet/sec?
- After how many seconds will it return to the ground when v0 = 96 feet/sec?
Homework Answers
s = –16t2 + v0t.
a.
s = 112 ft, v0 = 96 ft/s
Therefore,
112 = –16t2 + 96t.
16t2 - 96t + 112 = 0
Solvign the equation, we get
t = 4.41s and t = 1.58s
b.)
when at ground, s = 0
Therefore
–16t2 + 96t. = 0
16t(-t + 6) = 0
t = 0, and t = 6
Therefore, after 6 seconds, it will return to ground.
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