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anonymous
 one year ago
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
anonymous
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Set h(t) equal to 200 and solve for t

MrNood
 one year ago
Best ResponseYou've already chosen the best response.0and also notice by the way  on EARTH  gravity acceleration is 32 ft/s^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.016t² + 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 <

MrNood
 one year ago
Best ResponseYou've already chosen the best response.0the 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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