# Totally Lost (Free Fall Distance)

physicsvirgin

I have a conceptual problem more than just a homework problem this time. It regards free fall distance and the formula applied.

Okey dokey...the formula to find out how far an object falls from rest is:
d=1/2 gt^2

So I did some sample problems to get a hang of it...the problems consist of shooting an arrow straight up at 50m/s.

One problem is: Where is the arrow you shoot up at 50 m/s when it runs out of speed? Answer: 125m. I get this by multiplying 50(g) by 5(t) and dividing by 2.

Next problem: How high will the arrow be 7 seconds after being shot up at 50m/s? Answer: 105m. I got this by starting with the above answer, 125m because that is the distance traveled until the arrow stopped. To account for the 2 extra seconds, the arrow is falling at 10m/s so I subtracted 20 from 125 to get 105m.

My trouble is coming from the fact that I never used the ^2 portion of the formula to come to these conclusions. I simply multiplied acceleration and time and divided by 2. When was I supposed to square the time, and why did I come up with the correct answers without doing it???

Related Introductory Physics Homework Help News on Phys.org
You don't seem to understand the equation or concept.
The basic kinematic equation: $$x=v_0t+\frac{1}{2}at^2$$
The acceleration in this case is g which is $$-9.81 \frac{m}{s^2}$$

...One problem is: Where is the arrow you shoot up at 50 m/s when it runs out of speed? Answer: 125m. I get this by multiplying 50(g) by 5(t) and dividing by 2.

>> the constant 50 must not involve in the solution to get 125m and you use g= 10m/s^2 rather than 9.8m/s^2..
________________________________________________________________________
...My trouble is coming from the fact that I never used the ^2 portion of the formula to come to these conclusions. I simply multiplied acceleration and time and divided by 2. When was I supposed to square the time, and why did I come up with the correct answers without doing it???

>> it really return the same answer but not at all time, its just a coincidence
the exact solution is from the formula; y= 1/2g*t^2 (no initial velocity from the point the ball out of speed at t=5s)

you use g=10m/s^2

1. right: y= 1/2*10*2^2= 20m; 125-20= 105