Question

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

Answer 1

2.5?

Answer 2

Set h(t) equal to 200 and solve for t

Answer 3

and also notice by the way - on EARTH - gravity acceleration is -32 ft/s^2

Answer 4

-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 <-----

Answer 5

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