Vertical upward displacement equation. Time it takes to reach a certain height?

Question

Problem: an object is shot strait upward with an intial velocity of 6 m/s. Measure how long it takes for an object to reach the height of 1 meter?
Use the displacement equation to find the time at which the projectile would reach the height of 1 meter above its initial position.?
I know to solve this I need to change displacement equation to:
y = vot - 1/2 at^2
I'm just not sure how to plug in the numbers.

I think at= 20 but im not sure on how to put that into how long it will take to reach 1 meter?
Any help would be appreciated . Thank you..

Answer 1

initial velocity = 6 m/s
displacement = 1 meter
"g" = -9.8 m/sec^2


displacement = initial velocity * time - (1/2)(9.8)time^2
1 = 6t - 4.9t^2

4.9t^2 - 6t + 1 = 0

t = [ +6 +/- sqrt( (-6)^2 - (4)(4.9)(1) ) ] / (2)(4.9)
=[ 6 +/- sqrt( 36 - 19.6) ] / 9.8
t = 1.03 seconds or 0.20 seconds

The projectile will hit 1 meter twice, since eventually it will stop going upward and fall back to earth. But, you are looking for the first time it hits 1 meter, which is t = .20 seconds

Answer 2

I would rather suggest you to use a as 10 m/ sec², as it would simplify the equation

you get

1 = 6t - (1/2)5 t²

so u get t = 1/5 and 1

u get two possible answers because you reach that height twice...once when the object is going up and second time when it is falling back

Answer 3

y = vot - 1/2 at^2

set y = 1 meter, V0 = 6 meter/sec, a = 9.8 meter/sec^2
and solve for t.