A projectile is launched straight up with an initial velocity of 120 ft/s. If acceleration due to gravity is –16 ft/s2, after about how many seconds will the object reach a height of 200 ft? h(t) = at2 + vt + h0 1.0 s 1.7 s 2.5 s 8.0 s

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A projectile is launched straight up with an initial velocity of 120 ft/s. If acceleration due to gravity is –16 ft/s2, after about how many seconds will the object reach a height of 200 ft? h(t) = at2 + vt + h0 1.0 s 1.7 s 2.5 s 8.0 s

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2.5?
Set h(t) equal to 200 and solve for t
and also notice by the way - on EARTH - gravity acceleration is -32 ft/s^2

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Other answers:

-16t² + 120t = 200 -16t² + 120t - 200 = 0 d = √(120² - 4(-16)(-200) d = 40 t = (1/(2(-16))(-120 ± 40) t = (-1/32)(-120 ± 40) t = (-1/32)(-80) and (-1/32)(-160) t = 2.5 and 5 c) 2.5 s <-----
the equation you have written is not correct either - you have combined 2 errors to get a working answer. The correct equation for motion under gravity is \[h(t) = \frac{ at ^{2} }{ 2 } + vt +h _{0}\] you have missed out the factor of 1/2 in the equation and taken the acceleration to be 16 this is not correct - the equation as I wrote is true fro all values for constant acceleration. The equation YOU wrote is only true for earth's gravity

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