The acceleration of an object g = -9.8 m/s^2. From the top of a building of 245 m., an object is thrown upwards at an initial velocity of 12m/s

(a) What is the function for the velocity of the object?

I put vf^2 = Vi^2 + 2a(yf-yi)
vf^2 = 216.09 + 19.6 (yf - 245)

(b) Determine the function for the position of the object relative to the ground.

I don't know what to put for this one

(c) How long will it take for the object to reach maximum height?

I know that for maximum height velocity = 0 with the formula: Y = Yo + Vot+ 1/2at^2, but the problem is that when I try to isolate t, I still have the t from Vot which I don't know where to put, so I'm left with t = squ(2(Yo + Vot - Y))/a

(d) What is the velocity of the object once it reaches the ground?

For this one I'll use the same formula as for (a) but will my initial height be the maximum height reached from (c)?

my equation would be

s(t) = -4.9t^2 + 12t + 245

v(t) = -9.8t + 12

at the maximum height, v = 0
-9.8t + 12 = 0
t = 12/9.8 = 1.22 seconds

when it hits the ground, s(t) = 0
-4.9t^2 + 12t + 245 = 0
or
4.9t^2 - 12t - 245 = 0

using the formula ...
t = 8.4 or a negative

so at 8.4 seconds, v(8.4) = -9.8(8.4) + 12 = -70.32 m/sec

so for (a) you used Y = Yo + Vot+ 1/2at^2 so when you derive it, you get time for the other ones, got it. THanks a lot Reiny

(a) To find the function for the velocity of the object, you correctly used the equation vf^2 = Vi^2 + 2a(yf - yi), where vf is the final velocity, Vi is the initial velocity, a is the acceleration, yf is the final position, and yi is the initial position. However, you made a small mistake in your calculation. Let's re-calculate it correctly:

Given:
Vi = 12 m/s (initial velocity)
a = -9.8 m/s^2 (acceleration)

Let's assume the initial position (yi) is at the top of the building, so yi = 245m. The final position (yf) can take any value as we are finding a general function.

Using the formula:
vf^2 = Vi^2 + 2a(yf - yi)

Substituting the given values:
vf^2 = 12^2 + 2(-9.8)(yf - 245)
vf^2 = 144 - 19.6(yf - 245)
vf^2 = 144 - 19.6yf + 4802

Simplifying:
vf^2 = -19.6yf + 4946

Therefore, the function for the velocity of the object is:
v(y) = ±√(-19.6y + 4946) (± because there are two possible velocities - one when the object is going up, and one when it is going down)

(b) To determine the function for the position of the object relative to the ground, we need to integrate the velocity function obtained in part (a) with respect to y (position).

Integral of v(y) = ±√(-19.6y + 4946) dy

Integrating, we get:
y = -(5/49)(-19.6y + 4946)^(3/2) + C

Here, C is the constant of integration. To determine its value, we need an additional piece of information, like the initial position (yi) or the value of y at a certain time. Without this information, we can only provide a general function.

(c) To find the time it takes for the object to reach maximum height, we can use the fact that the velocity at maximum height is 0.

Using v(y) = ±√(-19.6y + 4946), we set this expression equal to 0 and solve for y:
0 = -19.6y + 4946

Simplifying, we find:
y = 4946/19.6 = 252.04 m

Now, we can use the formula you mentioned correctly: Y = Yo + Vot + 1/2at^2.

At maximum height, the object's final velocity (Vf) is 0, so we have:
0 = 12 - 9.8t

Solving for t, we get:
t = 12/9.8 = 1.224 s

Therefore, it takes approximately 1.224 seconds for the object to reach the maximum height.

(d) To find the velocity of the object once it reaches the ground, we need to find its final velocity when it hits the ground. Assuming the ground level as the reference (y = 0), we can find the time it takes for the object to reach the ground by setting y = 0 in the position function obtained in part (b).

y = -(5/49)(-19.6y + 4946)^(3/2) + C
0 = -(5/49)(-19.6(0) + 4946)^(3/2) + C

Simplifying, we find:
C = (5/49)(-19.6(0) + 4946)^(3/2)

Now, we can use this value of C to find the time (t) it takes for the object to reach the ground:

0 = -(5/49)(-19.6t + 4946)^(3/2) + (5/49)(-19.6(0) + 4946)^(3/2)

Simplifying, we find:
t = √(4946/19.6) = 10 s

Therefore, it takes approximately 10 seconds for the object to reach the ground, and its velocity at that moment will be calculated using the velocity function obtained in part (a) by substituting y = 0.

v(0) = ±√(-19.6(0) + 4946)
v(0) = ±√4946
v(0) = ±70.29 m/s

So, the velocity of the object once it reaches the ground is approximately ±70.29 m/s.