Calculating Drop Time to Reach 90 km/h: Confirm/Deny!

[KingNothing]

Can anyone confirm/deny that if an object is dropped (init. velocity=0) that it will take about 2.26 seconds to reach 90 km/h?

$$S=\frac{D}{T}$$
$$A=\frac{V_f-V_i}{T_f-T_i}$$
$$D=\frac{1}{2}gT^2$$
$$V_f^2=V_i^2+2a(X_f-X_i)$$
$$\overline{V}=\frac{V_f+V_i}{2}$$

 

[Jimmy]

Assuming no air resistance and an acceleration of 9.81 m/s^2, I get 2.55 seconds.

Or if the object falls for 2.26 seconds, it will have a velocity of 80 Km/hr.

 

[KingNothing]

Can you go through it for me? It's really confusing me as to why I keep getting what I do.

 

[Jimmy]

Time = Velocity/acceleration

The units I used were Seconds, m/s, m/s^2

I converted your value of 90 km/hr to m/s. Divide by 3.6 to convert km/hr to m/s. 90 km/hr = 25 m/s

I simply plugged in the values into the above formula.

T = 25 /9.81 = 2.55 seconds.

Your answer wans't really that far off. What values did you use?

 

[KingNothing]

Just to start off, we have to assume gravity to be 9.8 in our physics class, but I don't think that can attribute to the difference. Anyhoo, what is wrong with using:

$$V_f = V_i + \frac{1}{2} a t^2$$
If $$V_f$$=25 and $$\frac{1}{2}a$$=4.9 and $$V_i$$=0, then what should I come out with, or is this the wrong equation?

 

[Jimmy]

It's the wrong equation. $1/2 a t^2$ is an expression of distance traveled under a constant acceleration for a given length of time.

Since $V_i=0$, your equation simplifies to $V_f = 1/2 a t^2$ which is not correct. $D = 1/2 a t^2$ when intial distance and velocity equal 0. Otherwise,

$$D = D_i+V_i t + \frac{a t^2}{2}$$

the units on one side of the equation have to match the units on the other side. That's a good way to check for errors in an equation.

$$V_f = V_i + a t$$