Kinematics Velocity Help: Solving for Height with Gravity and Initial Velocity

[Dude22]

Homework Statement

I am not going to use the exact same problem, as that would be cheating as this is for school, but Instead i will use an example problem:

A can of pop is tossed directly upwards with an initial velocity of 4 m/s. How high will it go?

Homework Equations

I know that gravity is -9.8m/s/s so that is my acceleration

I know the initial velocity is 4m/s

and I know the final velocity is 0m/s

The Attempt at a Solution

What I instinctively do is simply, the initial velocity divided by the maceration which gets me 0.41 but I don't know if that is correct or if that would give me the time it took or the distance it traveled.

As this is not an actual problem from my homework, only an example problem to help me understand a concept, any help would be appreciated.

Thanks,
Josh

 

[haruspex]

I assume you meant acceleration. (Maceration is something to do with brewing.) Velocity has dimension LT^-1 (distance/time), accn has dimension LT^-2, so velocity/accn gives T.
Easiest way to get the height is by conservation of energy.

 

[rude man]

k.e + p.e. = constant

 

[kris2fer]

What kinematics formulas do you know that may be relevant to this problem?

 

[KClose1983]

ΔV= (V_f - V_i) = at

V_f = 0, V_i = 4, a = g = -9.8

(0 - 4) = (-9.8)t

- 4 = -9.8t

.41 = t

 

[agm1984]

vector wise, if you start somewhere and go up in the air and come back down the resultant will be 0 because you landed in the same spot you started from right?

consider that a secret hint

break it down into what you know:
a= -9.8m/s
d=?
t=?
V_o=4m/s
V_f= _____

can you get 3 of those to solve for a 4th one?

 

[FaraDazed]

Use: $v^2=u^2+2as$

where v is final velocity = 0, u is inital velocity = 4m/s, and a is -9.8m/s^2. s is the displacement which is what you are looking for.