Okay. An object falls at a rate of 32 feet per second per second, no?
Sometimes I can visualize this and it seems to make sense until the next time when it just turns my brain to mush. It’s one of those things that just won’t stay with me.
Can anyone suggest a way of thinking of this that might successfully implant itself in my old, cell depleted synapses?
How fast is an object falling after 5 seconds? After 10 seconds?
Nactually, it falls at 1/2 x g x (t-squared). On earth, g=32 feet.
after 1 sec., it’s fallen 16ft.
2 sec. = 64 ft.
3 sec. = 144 ft., etc.
At some point it will hit “terminal velocity”, its maximum speed. That varies by air pressure and the shape of the object.
Just multiply 16 times seconds-squared.
O man, I am not awake yet. I was calculating distance, not speed. Can’t remember the formula for instantaneous velocity off the top of my head.
I’m going to go away now, because I’ve been making too many dumb posts in the last 12 hours.
If you stated it more carefully, it might help you remember. Rather than saying it “falls at a rate of …”, it would be better to say “ACCELERATES at 32 feet per second per second”. That reminds you that it is the acceleration that is constant, which is the reason that it is expressed as a rate of speed (feet per second), per second, acceleration being the rate of change of speed.
Which means actual rate of speed is simple multiplication - 5 seconds, 32 * 5 = 160 ft/sec. 10 seconds, 32 * 10 = 320 ft/sec. You also note that that makes the units work out - multiplying (ft/sec)/sec times sec gives you ft/sec. All disregarding air resistance and other effects, of course. And your textbook will probably say ft/sec**2 rather than (ft/sec)/sec, but I wanted to motivate this while keeping symbolic manipulation to a minimum.
yojimboguy is telling you how to find out how far it will fall. I think the OP was asking about speed.
It may help to know that rate is an acceleration, not a speed. It means that the speed increases by 32 ft/s every second.
After 1 second of free-fall (neglecting the effects of drag), the object will be traveling at 32 ft/s, or about 21 mph. After 2 seconds, it will be moving at 64 ft/s., 160 ft/s in 5 seconds, 320 ft/s in 10 seconds, and so on.
This all ignores drag, which would slow down actual acceleration and eventually limit speed to some terminal velocity. A skydiver’s terminal velocity in the usual position is in the 160-175 ft/s range, for example.
I must learn to type faster. Ignore my post, yojimboguy and yabob both did a better and faster job than me.
Thanks all. I’m going to drum this into my head as I’m going to sleep next time and see if I wake up with a permanent understanding.
Ha!