SOLUTION: A projectile is launched from ground level with an initial velocity of v0 feet per second. Neglecting air resistance, its height in feet t seconds after launch is given by s=-16
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Question 1203587: A projectile is launched from ground level with an initial velocity of v0 feet per second. Neglecting air resistance, its height in feet t seconds after launch is given by s=-16t^2+v0t . Find the time(s) that the projectile will (a) reach a height of 288 ft and (b) return to the ground when v0=144 feet per second.
(b) The projectile returns to the ground after
enter your response here second(s).
Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!
Part (a)
s = -16t^2 + v0*t
s = -16t^2 + 144t
288 = -16t^2 + 144t
0 = -16t^2 + 144t - 288
-16t^2 + 144t - 288 = 0
We can then factor
-16t^2 + 144t - 288 = 0
-16(t^2 - 9t + 18) = 0
-16(t - 3)(t - 6) = 0
t-3 = 0 or t-6 = 0
t = 3 or t = 6
Or we can use the quadratic formula
Plugged in a = -16, b = 144, c = -288
or
or
or
Either way, the two answers for part (a) are t = 3 and t = 6
The projectile reaches 288 ft at t = 3 seconds when going upward.
Upon its downward trajectory, it gets back to a height 288 ft at t = 6 seconds.
The projectile is above 288 ft for the interval 3 < t < 6. Otherwise, the projectile is at 288 ft or lower.
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Part (b)
s = -16t^2 + 144t
-16t^2 + 144t = s
-16t^2 + 144t = 0
-16t(t - 9) = 0
-16t = 0 or t-9 = 0
t = 0/(-16) or t = 9
t = 0 or t = 9
The projectile starts on the ground, so it makes sense that t = 0 is one solution.
The other solution is t = 9 which is when the rocket returns to the ground again.
The flight time is 9 seconds.
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