vertical motion and velocity/position functions

[dalasTR]

my notes from my teacher for these types of questions are not helpful at all. every time i think i know what to do and i get an aswer, i am wrong. i will type the questions first and then show what i tried.

For the following questions, use the position function s(t) = -16t^2 + v(subscript 0)t + s(subscript0) for free-falling objects.
note: v(subscript 0)t should mean initial velocity and s(subscript0) would be initial height according to my teacher's notes

1. A silver dollar is dropped from the top of a building that is 1362 feet tall
(A) determine the position and velocity functions for the coin.
my work: based on the notes i thought the functions would be v(t) = -16t^2 and s(t)= -16t^2 + 1362
(B) determine the average velocity on the interval [1,2]
my work: i got the avg velocity to be -48 ft/sec because i plugged in the interval points and did [s(2)-s(1)]/[2-1]?
(C) find the instantaneous velocities when t=1 and t=2
my work: at t=1, -32 ft/sec and at t=2, -64 ft/sec
(D) find the time required for the coin to reach ground level
my work: i didnt know how to set up the equation
(E) find the velocity of the coin at impact
my work: again im not sure how to write the equation:


2. A ball is thrown straight down from the top of a 220ft building with an initial velocity of -22ft per/sec. What is its velocity after 3 seconds? After falling 108 feet?
my work: once more im not sure how to write the equation but this is what i tried:
-16t^2 - 22t + 220, then i plugged in 3 for t and got 10. and i had no idea how to do the second part

 

[Deleted member 4993]

my notes from my teacher for these types of questions are not helpful at all. every time i think i know what to do and i get an aswer, i am wrong. i will type the questions first and then show what i tried.

For the following questions, use the position function s(t) = -16t^2 + v(subscript 0)t + s(subscript0) for free-falling objects.
note: v(subscript 0)t should mean initial velocity and s(subscript0) would be initial height according to my teacher's notes

1. A silver dollar is dropped from the top of a building that is 1362 feet tall
(A) determine the position and velocity functions for the coin.
my work: based on the notes i thought the functions would be v(t) = -16t^2 and s(t)= -16t^2 + 1362

v(t) = u + at

in this case:

v(t) = -32 * t

(B) determine the average velocity on the interval [1,2]
my work: i got the avg velocity to be -48 ft/sec because i plugged in the interval points and did [s(2)-s(1)]/[2-1]?
(C) find the instantaneous velocities when t=1 and t=2
my work:at t=1, -32 ft/sec and at t=2, -64 ft/sec
(D) find the time required for the coin to reach ground level .....s(t)= -16t^2 + 1362 At ground level what is 's'?
my work: i didnt know how to set up the equation
(E) find the velocity of the coin at impact
my work: again im not sure how to write the equation:... v(t) = -32 * t .... use 't' from above.


2. A ball is thrown straight down from the top of a 220ft building with an initial velocity of -22ft per/sec. What is its velocity after 3 seconds? After falling 108 feet?
my work: once more im not sure how to write the equation but this is what i tried:
-16t^2 - 22t + 220, then i plugged in 3 for t and got 10. and i had no idea how to do the second part

 

[dalasTR]

so:
a: im still not sure if i understand. what is "u + at" ???
b: -48 ft/sec
c: -32 ft/sec at t=1 an -64 ft/sec at t=2
d: approx 9.226 seconds
e: i plug in 9.226? and get approx -295.242ft/sec


as for number two, any help?

 

[wjm11]

Re 1a):

v(t) = -16t^2

This is incorrect.

“v(t) = u + at” as Subhotosh stated. The “u” is just another way of writing “v(subscript 0)”. In other words, velocity equals “initial velocity” plus “acceleration times time”. Your position function is correct.

In this problem, initial velocity is zero because the object was “dropped.”

Acceleration (near the surface of the Earth) is –32 ft/sec^2, so the velocity function is

v(t) = 0 + -32t = -32t

Re 2):

2. A ball is thrown straight down from the top of a 220ft building with an initial velocity of -22ft per/sec. What is its velocity after 3 seconds? After falling 108 feet?
my work: once more im not sure how to write the equation but this is what i tried:
-16t^2 - 22t + 220, then i plugged in 3 for t and got 10.

You put the “-22” and the "220" in the correct places, but you failed to write the whole equation:

s(t) = -16t^2 - 22t + 220

What is “s(t)”? Is it velocity or position? Don’t just grab equations and plug number in. Be sure you have the right equation and know what you are solving for. When you “plugged in 3 for t and got 10”, what did that 10 represent? (The answer: at t = 3s, the ball is 10 feet above the ground.)

You need to use the velocity function. In this problem it’s

v(t) = -22 + -32t

Make sense?