Please Help!! Velocity Problem!?

Question

Near the earth's surface every particle experiences a constant acceleration of 32 ft/sec^2 directed toward the center of the earth; if "up" is considered the positive direction, then this is an acceleration of -32ft/sec^2. Consider a rock thrown straight up in the air from the ground. Let's denote time t=0 as the time when the rock reaches 12 feet up in the air at (upward) velocity of 4 feet per second. Fill in the following with numbers that differ from the correct value by no more than .1 units.

The rock left the ground (was thrown up in the air) at time t = ANSWER seconds, with initial velocity of ANSWER feet per second. It will fall back to the ground at time t = ANSWER second after reaching the maximum height of ANSWER feet at time t = ANSWER seconds.

Answer 1

a = -32
v = -32t + C When t= 0, v = 4 so C = 4
v= -32t+4
s = -16t^2 +4t + C when t=0 s = 12 so C = 12
s = -16t^2 +4t +12
t = -.875 seconds
vo = 25.4 ft/sec
.75 seconds
12 feet
.25 seconds

Answer 2

At time zero, the final velocity Vf = 4 ft/sec

This is also equal to the initial velocity Vi -32T where T is the time the rock was in the air (from the time of launch until it reached 4 ft./sec).

So

(1) Vi - 32T = 4
(2) Vi = 4 + 32T

Now the distance D that the rock travelled up to this point is equal to the average velocity multiplied by the time

(3) D = 12 = [(Vi + Vf) / 2] * T
Substitute for Vi using (2) and simplify. A quadratic equation is the result:

4T^2 + T - 3 = 0

T = 0.75

But the problem says that time zero is when the rock reached 4 ft per sec., so relative to that time, the time of the launch is minus 0.75 secs

Substitute in (3) to get Vi which equals 28 ft. sec

I won't do all the math, but you can now compute the maximum height (15.75 ft.), the total time until the rock reaches zero velocity (the apex of its flight (0.875 secs)) and I leave it to you to do the rest.

Answer 3

You know the y" (t) = -32. You can integrate once and use initial velocity to determine the constant of integration. Likewise another integration will give you the position and with the initial position you again can find the constant for that integration. The answers are contained in the following sets: a( {-.75,-1,33,1, 2}, b) {2, 24,28}...note the initial velocity Is 4 since initial means
t = 0 , c) {1/2,7/8,1, 8/7}, d) {11.5,11.75,12.25,12.5} ,
e) {-1.33,-0.75,1, 2}